模拟赛小结:2018 China Collegiate Programming Contest Final (CCPC-Final 2018)

比赛链接:传送门

跌跌撞撞6题摸银。

封榜后两题,把手上的题做完了还算舒服。就是罚时有点高。

开出了一道奇奇怪怪的题(K),然后ccpcf银应该比区域赛银要难吧,反正很开心qwq。



Problem A. Mischievous Problem Setter 00:14 (-2) Solved by Dancepted

良心签到题。WA2吃乳猪。

代码:

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl ‘\n‘
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == ‘-‘) fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-‘0‘, ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar(‘-‘), x = -x;
    do{++len; c[len] = x%10 + ‘0‘;} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

struct Node{
    int d, t;
    bool operator < (const Node& x) const {
        return d < x.d;
    }
}nodes[N];
int main() {
    fast;
    int T; cin >> T;
    for (int kase = 1; kase <= T; kase++) {
        int n, m; cin >> n >> m;
        for (int i = 1; i <= n; i++) {
            cin >> nodes[i].d;
        }
        for (int i = 1; i <= n; i++) {
            cin >> nodes[i].t;
        }
        sort(nodes+1, nodes+1+n);
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            if (nodes[i].t <= m) {
                m -= nodes[i].t;
                ans++;
            }
            else {
                break;
            }
        }
        cout << "Case " << kase << ": " << ans << endl;
    }
    return 0;
}

Problem L. Ultra Weak Goldbach‘s Conjecture  00:47(+) Solved by xk (miller rabin + 素数密度 + 哥德巴赫猜想)

根据素数密度为$log^{2}N$的结论,可以用米勒-拉宾的板子O(logn)判断大素数,暴力找出比n小的最大的一个大素数。

哥德巴赫猜想在小数据范围内成立,剩下部分如果是奇数就分成2 + 2 + 3 + 两个素数,如果是偶数就是2 + 2 + 2 + 两个素数。

(xk才是真正的数学选手,我连哥德巴赫猜想都不知道,就是打酱油的)

代码:$O(T × log^{3}N)$

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define INF 0x3f3f3f3f
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl ‘\n‘

using namespace std;
typedef long long ll;
typedef double db;

int random(int l, int r)
{
    return rand() % (r - l) + l;
}

ll fmul(ll a, ll b, ll mod)
{
    a %= mod;
    ll res = 0;
    for(;b;b>>=1) {
        if(b & 1) res = (res + a) % mod;
        a = (a + a) % mod;
    }
    return res;
}

ll fpow(ll a, ll b, ll mod)
{
    ll res = 1;
    for(;b;b>>=1) {
        if(b & 1) res = fmul(res, a, mod);
        a = fmul(a, a, mod);
    }
    return res;
}

bool witness(ll a, ll n, ll u, ll t)
{
    ll x0 = fpow(a, u, n), x1;
    for(int i = 1; i <= t; i++)
    {
        x1 = fmul(x0, x0, n);
        if(x1 == 1 && x0 != 1 && x0 != n - 1) return false;
        x0 = x1;
    }
    if(x1 != 1) return false;
    return true;
}

bool isprime(ll n, int times = 20)
{
    if(n == 2) return true;
    if(n < 2 || !(n & 1)) return false;
    ll u = n - 1, t = 0;
    while(u % 2 == 0) {
        t++;
        u>>=1;
    }
    while(times--)
    {
        ll a = random(1, n - 1);
        if(!witness(a, n, u, t)) return false;
    }
    return true;
}

int main()
{
    srand(time(0));
    fast;
    int T;
    cin >> T;
    for(int kase = 1; kase <= T; kase++)
    {
        cout << "Case " << kase << ": ";
        ll n;
        cin >> n;
        if(n < 12) {
            cout << "IMPOSSIBLE\n";
            continue;
        }
        for(ll i = n - 10; ; i--)
        {
            if(isprime(i)) {
                n -= i;
                cout << i;
                break;
            }
        }
        if(n & 1)
        {
            cout << " 2 2 3";
            n -= 7;
        }
        else
        {
            cout << " 2 2 2";
            n -= 6;
        }
        for(ll i = 2; i <= n / 2; i++)
        {
            if(isprime(i) && isprime(n - i))
            {
                cout << ‘ ‘ << i << ‘ ‘ << n - i << endl;
                break;
            }
        }
    }
}

Problem G. Pastoral Life in Stardew Valley  01:13 (+) Solved by Dancepted (平方和公式)

设$f_{n, m}$表示n × m的草地上放稻草人的方案数,则:

$f_{n, m} = \sum_{i=1}^{n-2} \sum_{j=1}^{m-2}(n-i+1) × (m-j+1) = \frac{(n-1)(n-2) × (m-1)(m-2)}{4}$

设$F_{n, m}$表示n × m的土地上的答案,则:

$F_{n, m} = \sum_{i=3}^{n}\sum_{j=3}^{m} (n-i+1)×(m-j+1)×f_{i, j}  $

$= \sum_{i=3}^{n}\sum_{j=3}^{m} (n-i+1)×(m-j+1)×\frac{1}{4}i(i-1) × i(i-1)$

$= \frac{1}{4} \sum_{i=3}^{n}(n-i+1)(i-1)(i-2)\sum_{j=3}^{m}(m-j+1)(j-1)(j-2)$

令$g_{x} = \frac{1}{2} \sum_{i=3}^{x}(x-i+1)(i-1)(i-2)$,则$F_{n, m} = g_{n} * g_{m}$。

考虑预处理$g_{x}$:

①:$g_{3} = 1$

②:若已知$g_{x} = \frac{1}{2} \sum_{i=3}^{x}(x-i+1)(i-1)(i-2)$,则:

$g_{x+1} = \frac{1}{2} \sum_{i=3}^{x+1}(x-i+1+1)(i-1)(i-2)$

$= \frac{1}{2} \sum_{i=3}^{x+1}(x-i+1)(i-1)(i-2) + \frac{1}{2}\sum_{i=3}^{x+1}(i-1)(i-2) $

令$h_{x} =  \frac{1}{2}\sum_{i=3}^{x}(i-1)(i-2) $,则:

$g_{x+1} = g_{x} + h_{x}$,其中,用平方和公式等差数列求和公式可以O(1)地计算$h_{x}$。

代码:O(T + N)

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl ‘\n‘
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == ‘-‘) fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-‘0‘, ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar(‘-‘), x = -x;
    do{++len; c[len] = x%10 + ‘0‘;} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

#define md 1000000007
ll mul(ll a, ll b) {
    return a * b % md;
}
ll add(ll a, ll b) {
    ll res = (a+b) % md;
    if (res < 0) res += md;
    return res;
}
ll fpow(ll a, ll p) {
    ll res = 1;
    for (; p; p >>= 1) {
        if (p & 1)
            res = mul(res, a);
        a = mul(a, a);
    }
    return res;
}

ll inv6, inv2;
ll g[N];
ll h(ll x) {
    ll res = 0;
    res = add(res, mul(mul(x, mul(x+1, 2*x+1)), inv6));
    res = add(res, mul(mul(x, x+1), inv2));
    res = mul(res, inv2);
    return res;
}
void init() {
    g[3] = 1;
    for (int i = 4; i < N; i++) {
        g[i] = add(g[i-1], h(i-2));
    }
}
int main() {
    fast;
    int T; cin >> T;
    inv2 = fpow(2, md-2);
    inv6 = fpow(6, md-2);
    init();
    for (int kase = 1; kase <= T; kase++) {
        int n, m; cin >> n >> m;
        ll ans = mul(g[n], g[m]);
        cout << "Case " << kase << ": " << ans << endl;
    }
    return 0;
}

Problem K. Mr. Panda and Kakin  02:36 (-2) Solved by Dancepted & xk (欧拉定理 逆元 素数密度)

根据欧拉定理的推论,$i^{a}$ mod n的循环节长度为$\phi(n)$,并且把n分解为$\sum_{p\in prime}p_{i}^{m_{i}}$后若$m_{i}$ <= 1,则$i^{a}$ mod n为纯循环(参考纯循环小数意会一下)。

那么只要能把$FLAG^{2^{30}+3}$凑成$FLAG^{1 mod \phi(n)}$就行了。

实际上$(x^{a})^{b} = x^{a×b}$,所以如果我们能求出$2^{30}+3$关于$phi(n)$的逆元,那么就有$(Flag^{2^{30}+3})^{逆元} = Flag^{1 mod \phi(n)} = Flag$。

而这个逆元是肯定存在的,因为$2^{30}+3$是一个质数,而且考虑到n的生成方式,n = p × q,phi(n) = (p-1)×(q-1)。而 p-1,q-1 < $2^{30}+3$,因此($2^{30}+3, \phi(n)$)= 1。

求$\phi(n)$的时候考虑素数密度,可以$O(log^{2}n)$暴力地找出n的两个素因子。

然后快速幂会爆long long,要用快速乘,然后这题的log又比较大,$log^{2}$会tle,所以要用O(1)的快速乘

PS:第一次写脑抽了以为$(x^{a})^{c} = x^{a+b}$,幸好没过样例。

PPS:这里吹爆jls在ccpc-camp讲的数论div2,听完之后碰到欧拉定理完全不虚,然后在comet oj的直播回放里就可以看(jls的盛世美颜)了。

代码:O(T×logn)

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl ‘\n‘
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;
typedef long double ldb;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == ‘-‘) fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-‘0‘, ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar(‘-‘), x = -x;
    do{++len; c[len] = x%10 + ‘0‘;} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll fmul(ll a, ll b, ll md) {
    a %= md, b %= md;
    ll c = (ldb) a * b / md;
    ll ans = a * b - c * md;
    if (ans < 0) ans += md;
    else if (ans >= md) ans -= md;
    return ans;
}
ll fpow(ll a, ll p, ll md) {
    ll res = 1;
    for (; p; p >>= 1) {
        if (p & 1)
            res = fmul(res, a, md);
        a = fmul(a, a, md);
    }
    return res;
}

ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (a == 0 && b == 0) return -1;
    if (b == 0) {x = 1, y = 0; return a;}
    ll d = exgcd(b, a%b, y, x);
    y -= a/b*x;
    return d;
}
ll mod_reverse(ll a, ll n) {
    ll x, y;
    ll d = exgcd(a, n, x, y);
    if (d == 1) return (x % n + n) % n;
    return -1;
}

int main() {
    // fast;
    int T; cin >> T;
    for (int kase = 1; kase <= T; kase++) {
        ll n, c; read(n, c);
        ll g = gcd(n, c);
        ll flag = 0, phin = 0;
        if (g == 1) {
            ll x = sqrt(n+0.5);
            if (x % 2 == 0)
                x--;
            for (ll i = x; i >= 0; i -= 2) {
                if (n % i == 0) {
                    phin = (i-1) * (n/i -1);
                    break;
                }
            }
        }
        else {
            phin = (g-1) * (n/g - 1);
        }
        ll p = mod_reverse((1<<30)+3, phin);
        flag = fpow(c, p, n);

        printf("Case %d: %I64d\n", kase, flag);
    }
    return 0;
}
/*
3
181857896263 167005790444
218128229323 156323229335
352308724847 218566715941
*/

Problem I. Cockroaches  04:19 (-1) Solved by Dancepted & lh & xk 

大概是个思维题吧。。。封榜20分钟才调出来qwq。(不过好像是第一次封榜后过题?)

能消灭的最多的小强数量只有两种情况。设小强数最多的行和列对应的小强数是r和c,那么能消灭最多的数量要么是r+c,要么是r+c-1。

然后遍历小强数最多的行(列)上的小强,统计能消灭r+c和r+c-1的方案数就行了。

具体的就是遍历小强数最多的行(列)上的小强的时候,看这些小强是否恰巧在小强数最多的列(行),如果在的话,说明激光中心在这个小强所在点上时,能消灭的数量是r+c-1而不是r+c。

若r+c的数量为0,那么用同样的方法再统计一下小强数次多的行(列)与小强数最多的列(行)对r+c-1的贡献就行了。

小强的坐标上限是1e9,要离散化一下。

特别地:依次最多消灭小强数为2的时候要特判一下,防止在两个不同点消灭了两个相同小强。

代码:O(T×nlogn)

#include <iostream>
#include <cmath>
#include <map>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <iomanip>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 200005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl ‘\n‘
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == ‘-‘) fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-‘0‘, ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar(‘-‘), x = -x;
    do{++len; c[len] = x%10 + ‘0‘;} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }

int n;
vector<int> vals;
map<int, int> id;
// int id[N<<1];
struct Node{
    int r, c;
}ns[N];
vector <Node> vc[N], vr[N];
int cntr1 = -1, cntr2 = -1, lenr1 = -1, lenr2 = -1;
int cntc1 = -1, cntc2 = -1, lenc1 = -1, lenc2 = -1;
void init() {
    sort(vals.begin(), vals.end());
    vals.erase(unique(vals.begin(), vals.end()), vals.end());
    for (int i = 0; i < sz(vals); i++) {
        id[vals[i]] = i;
        vc[i].clear();
        vr[i].clear();
    }

    for (int i = 1; i <= n; i++) {
        int idr = id[ns[i].r], idc = id[ns[i].c];
        vr[idr].push_back(ns[i]);
        vc[idc].push_back(ns[i]);
    }
    cntr1 = -1, cntr2 = -1, lenr1 = -1, lenr2 = -1;
    cntc1 = -1, cntc2 = -1, lenc1 = -1, lenc2 = -1;
    for (int i = 0; i < sz(vals); i++) {
        if (sz(vr[i]) > lenr1) {
            lenr2 = lenr1;
            cntr2 = cntr1;
            lenr1 = sz(vr[i]);
            cntr1 = 1;
        }
        else if (sz(vr[i]) == lenr1) {
            cntr1++;
        }
        else if (sz(vr[i]) > lenr2) {
            lenr2 = sz(vr[i]);
            cntr2 = 1;
        }
        else if (sz(vr[i]) == lenr2) {
            cntr2++;
        }

        if (sz(vc[i]) > lenc1) {
            lenc2 = lenc1;
            cntc2 = cntc1;
            lenc1 = sz(vc[i]);
            cntc1 = 1;
        }
        else if (sz(vc[i]) == lenc1) {
            cntc1++;
        }
        else if (sz(vc[i]) > lenc2) {
            lenc2 = sz(vc[i]);
            cntc2 = 1;
        }
        else if (sz(vc[i]) == lenc2) {
            cntc2++;
        }
    }
}

int main() {
    fast;
    int T; cin >> T;
    for (int kase = 1; kase <= T; kase++) {
        cin >> n;
        id.clear();
        vals.clear();
        for (int i = 1; i <= n; i++) {
            read(ns[i].r, ns[i].c);
            vals.push_back(ns[i].r);
            vals.push_back(ns[i].c);
        }
        init();

        ll ans1 = lenc1 + lenr1, cnt1 = 0;
        ll ans2 = lenc1 + lenr1 - 1, cnt2 = 0;
        for (int i = 0; i < sz(vals); i++) {
            if (sz(vc[i]) == lenc1) {
                cnt1 += cntr1;
                if (lenr2 == lenr1 - 1) {
                    cnt2 += cntr2;
                }
                for (Node &tmp : vc[i]) {
                    if (sz(vr[id[tmp.r]]) == lenr1) {
                        // share same point
                        cnt1--;
                        cnt2++;
                    }
                    else if (lenr2 == lenr1 - 1 && sz(vr[id[tmp.r]]) == lenr2) {
                        cnt2--;
                    }
                }
            }
            else if (lenc2 == lenc1 - 1 && sz(vc[i]) == lenc2) {
                cnt2 += cntr1;
                for (Node &tmp: vc[i]) {
                    if (sz(vr[id[tmp.r]]) == lenr1) {
                        // share same point
                        cnt2--;
                    }
                }
            }
        }

        ll ans = 0, cnt = 0;
        if (cnt1 > 0) {
            ans = ans1, cnt = cnt1;
        }
        else {
            ans = ans2, cnt = cnt2;
        }
        if (ans == 2) {
            cnt = 1LL * n * (n-1) / 2;
        }
        printf("Case %d: %I64d %I64d\n", kase, ans, cnt);
    }
    return 0;
}

Problem B. Balance of the Force 04:35(+) Solved by lh(贪心)

不能放在同一边的两个人连一条边,如果得到的图中有奇数环,则不可能。

然后枚举最小的能力值。贪心地寻找最小的最大值。

枚举下一个最小的能力值时,仅有当前最小能力值所在的环,和下一个最小能力值所在的环对应的能力值要更新,所以整个贪心可以是O(N)的。

代码:O(T×N)

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <cstring>
#define N 200005
#define INF 0x3f3f3f3f
#define fi first
#define se second

using namespace std;
typedef pair<int,int> pii;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == ‘-‘) fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-‘0‘, ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
template <typename T>
inline void write(T x) {
    int len = 0; char c[21]; if (x < 0) putchar(‘-‘), x = -x;
    do{++len; c[len] = x%10 + ‘0‘;} while (x /= 10);
    for (int i = len; i >= 1; i--) putchar(c[i]);
}
template <typename T, typename... Args>
inline void write(T x, Args ... args) { write(x), write(args...); }
int T;
int n, m, ednum, top, col[N];
int w[N][2];
struct node
{
    int be, w;
    bool operator<(const node &other)const
    {
        return w < other.w;
    }
} s[N << 1];
struct Unite
{
    int cur, maxn[2], minx[2];
} st[N];
int hed[N << 1], nxt[N << 1], to[N << 1];
void add(int u, int v)
{
    to[++ednum] = v;
    nxt[ednum] = hed[u], hed[u] = ednum;
}
bool dfs(int v)
{
    st[top].maxn[0] = max(st[top].maxn[0], w[v][col[v]]), st[top].minx[0] = min(st[top].minx[0], w[v][col[v]]);
    st[top].maxn[1] = max(st[top].maxn[1], w[v][col[v] ^ 1]), st[top].minx[1] = min(st[top].minx[1], w[v][col[v] ^ 1]);
    for (int i = hed[v]; i; i = nxt[i])
    {
        int u = to[i];
        if (col[u] == col[v])
            return false;
        if (col[u] != -1)
            continue;
        col[u] = 1 ^ col[v];
        if (dfs(u) == false)
            return false;
    }
    return true;
}
int save[N], ansmax;
bool reduce()
{
    while (top)
    {
        int id = save[top];
        if (st[id].cur == 1)
            return false;
        ansmax = max(ansmax, st[id].maxn[1]), st[id].cur = 1, --top;
    }
    return true;
}
int main() {
    read(T);
    int u, v, cnt;
    int casecnt = 0;
    while (T--)
    {
        ++casecnt;
        read(n, m), ednum = top = 0, memset(hed, 0, sizeof(int) * (n + 1)), memset(col, -1, sizeof(int) * (n + 1));
        for (int i = 1;i <= m; ++i)
            read(u, v), add(u, v), add(v, u);
        for (int i = 1;i <= n; ++i)
            read(w[i][0], w[i][1]);
        bool flag = true;
        ansmax = 0, top = 0, cnt = 0;
        int ans = INF;
        for (int i = 1;i <= n; ++i)
        {
            if (col[i] != -1) continue;
            ++top, st[top].cur = 0, st[top].maxn[0] = st[top].maxn[1] = 0;
            st[top].minx[0] = st[top].minx[1] = INF, col[i] = 0, flag &= dfs(i);
            if (flag == false)
                break;
            if (st[top].minx[0] > st[top].minx[1])
                swap(st[top].minx[0], st[top].minx[1]), swap(st[top].maxn[0], st[top].maxn[1]);
            if (st[top].maxn[0] >= st[top].maxn[1])
                st[top].cur = 1;
            else
                s[++cnt] = node{top, st[top].minx[0]};
            s[++cnt] = node{top, st[top].minx[1]};
            ansmax = max(ansmax, st[top].maxn[st[top].cur]);
        }
        printf("Case %d: ", casecnt);
        if (flag == false)
        {
            puts("IMPOSSIBLE");
            continue;
        }
        sort(s + 1, s + 1 + cnt);
        int i = 1;
        top = 0;
        while (flag && i <= cnt)
        {
            ans = min(ans, ansmax - s[i].w);
            save[++top] = s[i].be;
            if (s[i].w != s[i + 1].w)
            {
                flag &= reduce();
                if (!flag)
                    break;
            }
            ++i;
        }
        write(ans), putchar(‘\n‘);
    }
    return 0;
}

还有不到一周就是CCPC-Final了,这周每两天一套题,冲鸭。



总结:

浮躁的菜逼选手贡献全部罚时。

原文地址:https://www.cnblogs.com/Lubixiaosi-Zhaocao/p/11831096.html

时间: 2024-10-06 04:28:36

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