ComWin’ round 11部分题解

https://vjudge.net/contest/325913#overview

A.Threehouses

题意：一直二维平面上的$n$个点中，前$e$个点落在小岛周围，并且有$p$条边已经连接，问最少花费使得所有点都可以通过一些边到达小岛，两点之间建边的花费为两点间的欧式距离。

思路：根据$kruskal$求最小生成树的方法将前$e$个点合并起来，再将已有$p$条边的两点合并，之后做一次$kruskal$把所有点合并即可。

#include<bits/stdc++.h>

using namespace std;
const int maxn = 1000 + 10;

int fa[maxn];
double x[maxn], y[maxn], dist[maxn][maxn];
int tot;
struct node {
    int u, v;
    double val;
    bool operator < (const node &rhs) const {
        return val < rhs.val;
    }
}edge[maxn * maxn];

int fr(int x) {
    if(fa[x] == x) return x;
    return fa[x] = fr(fa[x]);
}
void uni(int x, int y) {
    x = fr(x), y = fr(y);
    if(x != y) fa[x] = y;
}
void init(int n) {
    tot = 0;
    for(int i = 0; i <= n; i++) {
        fa[i] = i;
    }
}
double cal(int i, int j) {
    double ret = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
    return ret;
}
int main() {
    int n, e, p;
    scanf("%d%d%d", &n, &e, &p);
    init(n + 1);
    for(int i = 1; i <= n; i++) {
        scanf("%lf%lf", &x[i], &y[i]);
    }
    for(int i = 1; i <= n; i++) {
        for(int j = i + 1; j <= n; j++) {
            dist[i][j] = dist[j][i] = cal(i, j);
            node now;
            now.u = i, now.v = j, now.val = dist[i][j];
            edge[tot++] = now;
        }
    }
    sort(edge, edge + tot);
    for(int i = 1; i <= e; i++) {
        uni(0, i);
    }
    double ans = 0.0;
    for(int i = 0; i < p; i++) {
        int u, v;
        scanf("%d%d", &u, &v);
        uni(u, v);
    }
    for(int i = 0; i < tot; i++) {
        node now = edge[i];
        int fu = fr(now.u), fv = fr(now.v);
        if(fu != fv) {
            ans += now.val;
            fa[fu] = fv;
        }
    }
    printf("%.6f\n", ans);
    return 0;
}

C.Cops ans Robbers

题意：给定图中只有字母点可以设立障碍，问让小偷所在的位置$B$无法达到边界所需设立障碍的最小花费。

思路：先说做法。考虑最小割，对每个点拆点，让$u$到$u^{‘}$的流量为该点对应的值，如果该点不能设立障碍则为$inf$，然后对一个点$u$的相邻点连一条$u^{‘}$到$v$流量为$inf$的边，如果越界那么就连向汇点$T$，流量依旧为$inf$。这样做使得每个格点所在的路径流向$T$的最小割只会被其上的价值最小的点所限制，由于建图时所有相邻格点都有连边，所以从源点出发时不会漏边。

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 1000 + 5;
const int inf = 0x3f3f3f3f;
const LL linf = 0x3f3f3f3f3f3f3f3f;

int n, m, c, T;
LL val[26];
char mp[N][N];
struct Dinic {
    static const int maxn = 1e6 + 5;
    static const int maxm = 4e6 + 5;

    struct Edge {
        int u, v, next;
        LL flow, cap;
    } edge[maxm];

    int head[maxn], level[maxn], cur[maxn], eg;

    void addedge(int u, int v, LL cap) {
        edge[eg] = {u, v, head[u], 0, cap}, head[u] = eg++;
        edge[eg] = {v, u, head[v], 0, 0}, head[v] = eg++;
    }

    void init() {
        eg = 0;
        memset(head, -1, sizeof head);
    }

    bool makeLevel(int s, int t, int n) {
        for(int i = 0; i < n; i++) level[i] = 0, cur[i] = head[i];
        queue<int> q; q.push(s);
        level[s] = 1;
        while(!q.empty()) {
            int u = q.front();
            q.pop();
            for(int i = head[u]; ~i; i = edge[i].next) {
                Edge &e = edge[i];
                if(e.flow < e.cap && level[e.v] == 0) {
                    level[e.v] = level[u] + 1;
                    if(e.v == t) return 1;
                    q.push(e.v);
                }
            }
        }
        return 0;
    }

    LL findpath(int s, int t, LL limit = linf) {
        if(s == t || limit == 0) return limit;
        for(int i = cur[s]; ~i; i = edge[i].next) {
            cur[edge[i].u] = i;
            Edge &e = edge[i], &rev = edge[i^1];
            if(e.flow < e.cap && level[e.v] == level[s] + 1) {
                LL flow = findpath(e.v, t, min(limit, e.cap - e.flow));
                if(flow > 0) {
                    e.flow += flow;
                    rev.flow -= flow;
                    return flow;
                }
            }
        }
        return 0;
    }

    LL max_flow(int s, int t, int n) {
        LL res = 0;
        while(makeLevel(s, t, n)) {
            LL flow;
            while((flow = findpath(s, t)) > 0) {
                if(res >= linf) return res;
                res += flow;
            }
        }
        return res;
    }
} di;

int id(int r, int c) {
    if(r < 1 || r > n || c < 1 || c > m) return T;
    return (r - 1) * m + c;
}

LL getVal(int r, int c) {
    if(r < 1 || r > n || c < 1 || c > m) return linf;
    if(mp[r][c] == ‘.‘ || mp[r][c] == ‘B‘) return linf;
    else return val[mp[r][c] - ‘a‘];
}

int main() {
    scanf("%d%d%d", &m, &n, &c);
    for(int i = 1; i <= n; i++) scanf("%s", mp[i] + 1);
    for(int i = 0; i < c; i++) scanf("%lld", &val[i]);
    T = 2*n*m + 5;
    di.init();
    di.addedge(id(1, 1) + n*m, T, linf);
    di.addedge(id(1, m) + n*m, T, linf);
    di.addedge(id(n, 1) + n*m, T, linf);
    di.addedge(id(n, m) + n*m, T, linf);
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            di.addedge(id(i, j), id(i, j) + n*m, getVal(i, j));
            if((i == 1 || i == n) && (j == 1 || j == m)) continue;
            di.addedge(id(i, j) + n*m, id(i+1, j), linf);
            di.addedge(id(i, j) + n*m, id(i-1, j), linf);
            di.addedge(id(i, j) + n*m, id(i, j+1), linf);
            di.addedge(id(i, j) + n*m, id(i, j-1), linf);
        }
    }
    LL ans = 0;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            if(mp[i][j] == ‘B‘) {
                ans = di.max_flow(id(i, j), T, T+10);
            }
            if(ans) break;
        }
        if(ans) break;
    }
    printf("%lld\n", ans >= linf ? -1 : ans);
    return 0;
}
/*
10 10 1
..........
..........
..........
...a.a....
..a.a.a...
.a..B.a...
..aaaaa...
..........
..........
..........
1
*/

E.Coprime Integers

题意：给定$a,b,c,d$，求$\sum_{i=a}^{b}\sum_{j=c}^{d}[gcd(i,j)=1]$，$[t=1]$表示$t$等于$1$时的值为$1$，否则为$0$。

思路：容斥原理+莫比乌斯反演，考虑$solve(n,m)=\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=1]$，可以发现$ans=solve(b,d)-solve(a-1,d)-solve(c-1,b)+solve(a-1,c-1)$。接着考虑如何计算$solve(n,m)$

$\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=1]$

$=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{t|i,t|j}u(t)$

$=\sum_{t=1}^{min(n,m)}u(t)\sum_{i=1}^{\left \lfloor \frac{n}{t} \right \rfloor}\sum_{j=1}^{\left \lfloor \frac{m}{t} \right \rfloor}1$

$=\sum_{t=1}^{min(n,m)}u(t)\left \lfloor \frac{n}{t} \right \rfloor\left \lfloor \frac{m}{t} \right \rfloor$

预处理莫比乌斯函数前缀和之后整除分块即可在$O(\sqrt n)$复杂度内求出一次$solve$。

#include<bits/stdc++.h>

using namespace std;
typedef long long LL;
const int N = 1e7 + 5;

int p[N / 10], cnt, mu[N];
bool tag[N];

void getPrime() {
    mu[1] = 1;
    for(int i = 2; i < N; i++) {
        if(!tag[i]) {
            mu[i] = -1;
            p[cnt++] = i;
        }
        for(int j = 0; j < cnt && 1LL * p[j] * i < N; j++) {
            tag[i * p[j]] = 1;
            if(i % p[j] == 0) {
                mu[i * p[j]] = 0;
                break;
            }
            mu[i * p[j]] = -mu[i];
        }
    }
    for(int i = 1; i < N; i++) mu[i] += mu[i - 1];
}

LL solve(LL n, LL m) {
    LL res = 0;
    for(LL l = 1, r; l <= min(n, m); l = r + 1) {
        r = min(n / (n / l), m / (m / l));
        res += (mu[r] - mu[l - 1]) * (n / l) * (m / l);
    }
    return res;
}

int main() {
    getPrime();
    LL a, b, c, d;
    scanf("%lld%lld%lld%lld", &a, &b, &c, &d);
    printf("%lld\n", solve(b, d) - solve(a - 1, d) - solve(c - 1, b) + solve(a - 1, c - 1));
    return 0;
}

H.Heir‘s Dilemma

题意：求$L$到$H$之间满足位数是$6$位，且没有某一位是$0$，且能被每一位的数字整除的数的个数。

思路：数据范围很小，暴力枚举每个数$check$是否可行即可。

#include <bits/stdc++.h>

using namespace std;

bool ck(int x) {
    bool vis[10];
    for(int i = 0; i < 10; i++) vis[i] = false;
    int xx = x, xxx = x, xxxx = x;
    while(xxx) {
        if(xxx % 10 == 0) return false;
        xxx /= 10;
    }
    while(xxxx) {
        if(vis[xxxx % 10] == true) return false;
        vis[xxxx % 10] = true;
        xxxx /= 10;
    }
    while(xx) {
        int now = xx % 10;
        if(x % now != 0) return false;
        xx /= 10;
    }
    return true;
}

int main() {
    int cnt = 0, a, b;
    cin >> a >> b;
    for(int i = a; i <= b; i++) cnt += (int)ck(i);
    cout << cnt << endl;
}

原文地址：https://www.cnblogs.com/DuskOB/p/11517554.html