湖南多校对抗5.24

据说A,B,C题都比较水这里就不放代码了

D:Facility Locations

然而D题是一个脑经急转弯的题：有m行，n列，每个位置有可能为0，也可能不为0，问最多选K行是不是可以使得每一列都至少有一个0，其中代价c有个约束条件：These costs satisfy a locality property: for two clients j and j’ and two facilities i and i’, we have cij ≤ ci’j + ci’j’ + cij’ .

一看特别像是重复覆盖的模板，然而稍加分析就会发现由于上面约束的存在，那么任意两行之间的0的摆放要么完全相同，要么完全错位，所以最后只要找出任意k个完全错开的行，判断是不是每一列都有一个0就可以了= =

 1 #include <map>
 2 #include <set>
 3 #include <stack>
 4 #include <queue>
 5 #include <cmath>
 6 #include <ctime>
 7 #include <vector>
 8 #include <cstdio>
 9 #include <cctype>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 #define INF 0x3f3f3f3f
16 #define inf (-((LL)1<<40))
17 #define lson k<<1, L, (L + R)>>1
18 #define rson k<<1|1,  ((L + R)>>1) + 1, R
19 #define mem0(a) memset(a,0,sizeof(a))
20 #define mem1(a) memset(a,-1,sizeof(a))
21 #define mem(a, b) memset(a, b, sizeof(a))
22 #define FIN freopen("in.txt", "r", stdin)
23 #define FOUT freopen("out.txt", "w", stdout)
24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
25 #define dec(i, a, b) for(int i = a; i >= b; i --)
26
27 template<class T> T CMP_MIN(T a, T b) { return a < b; }
28 template<class T> T CMP_MAX(T a, T b) { return a > b; }
29 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
30 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
33
34 //typedef __int64 LL;
35 typedef long long LL;
36 const int MAXN = 51000;
37 const int MAXM = 110000;
38 const double eps = 1e-4;
39 LL MOD = 1000000007;
40
41 int n, m, k, x;
42 bool vis[110];
43
44 int main()
45 {
46     while(~scanf("%d %d %d", &n, &m, &k)) {
47         mem0(vis);
48         int time = 0, col = 0;
49         rep (i, 1, n) {
50             int add = 0;
51             rep (j, 1, m) {
52                 scanf("%d", &x);
53                 if(x || vis[j]) continue;
54                 if(!add) time ++;
55                 add = 1;
56                 vis[j] = 1, col ++;
57             }
58         }
59         puts(col == m && time <= k ? "yes" : "no");
60     }
61     return 0;
62 }

E：Repeated Substrings 据说是一个后缀数组的最长公共前缀（LCP）的应用，

赛后又自己写了一遍后缀数组（其实是把模板打了一遍，模板见链接），终于还是把这题A了

题解见链接

  1 #include <map>
  2 #include <set>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <vector>
  8 #include <cstdio>
  9 #include <cctype>
 10 #include <cstring>
 11 #include <cstdlib>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 #define INF 0x3f3f3f3f
 16 #define inf (-((LL)1<<40))
 17 #define lson k<<1, L, (L + R)>>1
 18 #define rson k<<1|1,  ((L + R)>>1) + 1, R
 19 #define mem0(a) memset(a,0,sizeof(a))
 20 #define mem1(a) memset(a,-1,sizeof(a))
 21 #define mem(a, b) memset(a, b, sizeof(a))
 22 #define FIN freopen("in.txt", "r", stdin)
 23 #define FOUT freopen("out.txt", "w", stdout)
 24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 25 #define dec(i, a, b) for(int i = a; i >= b; i --)
 26
 27 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 28 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
 33
 34 //typedef __int64 LL;
 35 typedef long long LL;
 36 const int MAXN = 110000;
 37 const int MAXM = 110000;
 38 const double eps = 1e-4;
 39 LL MOD = 1000000007;
 40
 41 struct SufArray {
 42     char s[MAXN];
 43     int sa[MAXN], t[MAXN], t2[MAXN], c[MAXN], n, m;
 44     int rnk[MAXN], height[MAXN];
 45     int mi[MAXN][20], idxK[MAXN];
 46
 47     void init() {
 48         mem0(s);
 49         mem0(height);
 50     }
 51     void read_str() {
 52         gets(s);
 53         m = 128;
 54         n = strlen(s);
 55         s[n++] = ‘ ‘;
 56     }
 57     void build_sa() {
 58         int *x = t, *y = t2;
 59         rep (i, 0, m - 1) c[i] = 0;
 60         rep (i, 0, n - 1) c[x[i] = s[i]] ++;
 61         rep (i, 1, m - 1) c[i] += c[i - 1];
 62         dec (i, n - 1, 0) sa[--c[x[i]]] = i;
 63         for(int k = 1; k <= n; k <<= 1) {
 64             int p = 0;
 65             rep (i, n - k, n - 1) y[p++] = i;
 66             rep (i, 0, n - 1) if(sa[i] >= k) y[p++] = sa[i] - k;
 67             rep (i, 0, m - 1) c[i] = 0;
 68             rep (i, 0, n - 1) c[x[y[i]]] ++;
 69             rep (i, 0, m - 1) c[i] += c[i - 1];
 70             dec (i, n - 1, 0) sa[--c[x[y[i]]]] = y[i];
 71             swap(x, y);
 72             p = 1;
 73             x[sa[0]] = 0;
 74             rep (i, 1, n - 1) {
 75                 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++;
 76             }
 77             if(p >= n) break;
 78             m = p;
 79         }
 80     }
 81     void get_height() {
 82         int k = 0;
 83         rep (i, 0, n - 1) rnk[sa[i]] = i;
 84         rep (i, 0, n - 1) {
 85             if(k) k --;
 86             int j = sa[rnk[i] - 1];
 87             while(s[i + k] == s[j + k]) k ++;
 88             height[rnk[i]] = k;
 89         }
 90     }
 91     void rmq_init(int *a, int n) {
 92         rep (i, 0, n - 1) mi[i][0] = a[i];
 93         for(int j = 1; (1 << j) <= n; j ++) {
 94             for(int i = 0; i + (1<<j) - 1 < n; i ++) {
 95                 mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]);
 96             }
 97         }
 98         rep (len, 1, n) {
 99             idxK[len] = 0;
100             while((1 << (idxK[len] + 1)) <= len) idxK[len] ++;
101         }
102     }
103     int rmq_min(int l, int r) {
104         int len = r - l + 1, k = idxK[len];
105         return min(mi[l][k], mi[r - (1 << k) + 1][k]);
106     }
107     void lcp_init() {
108         get_height();
109         rmq_init(height, n);
110     }
111     int get_lcp(int a, int b) {
112         if(a == b) return n - a - 1;
113         return rmq_min(min(rnk[a], rnk[b]) + 1, max(rnk[a], rnk[b]));
114     }
115     void solve() {
116         get_height();
117         LL ans = 0, pre = 0;
118         rep (i, 1, n - 1) {
119             if(height[i] > pre) ans += height[i] - pre;
120             pre = height[i];
121         }
122         cout << ans << endl;
123     }
124 };
125
126 int T;
127 SufArray sa;
128
129 int main()
130 {
131     while(~scanf("%d%*c", &T)) while(T--){
132         sa.init();
133         sa.read_str();
134         sa.build_sa();
135         sa.solve();
136     }
137     return 0;
138 }
139 /**************************************************************
140     Problem: 1632
141     User: csust_Rush
142     Language: C++
143     Result: Accepted
144     Time:880 ms
145     Memory:13192 kb
146 ****************************************************************/

F：Landline Telephone Network题意就是求一颗最小生成树，但是任意两个点之间的唯一路径上不能通过任何一个 坏点  ，题目告诉你哪些是坏点

其实就是先不管坏点，求一遍最小生成树，然后把坏点加进去（注意细节地方）

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <algorithm>
 7 #include <cstring>
 8 #define mem0(a) memset(a, 0, sizeof(a))
 9 #define rep(i,a, b) for(int i = a; i <= b; i ++)
10 using namespace std;
11
12 int fa[1100];
13 struct Edge {
14     int u, v, w;
15     Edge(int _u = 0, int _v = 0, int _w = 0) {
16         u = _u; v = _v; w = _w;
17     }
18     bool operator < (const Edge &A) const {
19         return w < A.w;
20     }
21 }e[110000];
22 int n, m, p, x, b[1100], u, v, w;
23 int t[1100];
24
25 int findFa(int x) {
26     return x == fa[x] ? x : fa[x] = findFa(fa[x]);
27 }
28
29 int main()
30 {
31     //freopen("in.txt", "r", stdin);
32     while(~scanf("%d %d %d", &n, &m, &p) && n) {
33         mem0(b); mem0(t);
34         rep (i, 1, n) fa[i] = i;
35         rep (i, 1, p) scanf("%d", &x), b[x] = 1;
36         rep (i, 1, m) {
37             scanf("%d %d %d", &u, &v, &w);
38             e[i] = Edge(u, v, w);
39         }
40         if(n == 2) {
41             cout << w << endl;
42             continue;
43         }
44         sort(e + 1, e + m + 1);
45         int ans = 0;
46         rep (i, 1, m) {
47             u = e[i].u, v = e[i].v, w = e[i].w;
48             if(b[u] || b[v]) continue;
49             int x = findFa(u), y = findFa(v);
50             if(x != y) {
51                 ans += w;
52                 fa[x] = y;
53             }
54         }
55         int imposs = 0;
56         rep (i, 1, m) {
57             u = e[i].u, v = e[i].v, w = e[i].w;
58             if(!(b[u] ^ b[v])) continue;
59             t[u]++; t[v] ++;
60             if(b[u] && t[u] >= 2) continue;
61             if(b[v] && t[v] >= 2) continue;
62             int x = findFa(u), y = findFa(v);
63             if(x != y) {
64                 ans += w;
65                 fa[x] = y;
66             }
67         }
68         int cnt = 0;
69         rep (i, 1, n) if(findFa(fa[i]) == i) {cnt ++; }
70         if(imposs || cnt > 1) puts("impossible");
71         else cout << ans << endl;
72     }
73     return 0;
74 }
75
76 /**************************************************************
77     Problem: 1633
78     User: csust_Rush
79     Language: C++
80     Result: Accepted
81     Time:84 ms
82     Memory:2792 kb
83 ****************************************************************/

GAquarium Tank简单计算几何

  1 #include <map>
  2 #include <set>
  3 #include <stack>
  4 #include <queue>
  5 #include <cmath>
  6 #include <ctime>
  7 #include <vector>
  8 #include <cstdio>
  9 #include <cctype>
 10 #include <cstring>
 11 #include <cstdlib>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 #define INF 0x3f3f3f3f
 16 #define inf (-((LL)1<<40))
 17 #define lson k<<1, L, (L + R)>>1
 18 #define rson k<<1|1,  ((L + R)>>1) + 1, R
 19 #define mem0(a) memset(a,0,sizeof(a))
 20 #define mem1(a) memset(a,-1,sizeof(a))
 21 #define mem(a, b) memset(a, b, sizeof(a))
 22 #define FIN freopen("in.txt", "r", stdin)
 23 #define FOUT freopen("out.txt", "w", stdout)
 24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
 25 #define dec(i, a, b) for(int i = a; i >= b; i --)
 26
 27 template<class T> T CMP_MIN(T a, T b) { return a < b; }
 28 template<class T> T CMP_MAX(T a, T b) { return a > b; }
 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
 33
 34 //typedef __int64 LL;
 35 typedef long long LL;
 36 const int MAXN = 110000;
 37 const int MAXM = 110000;
 38 const double eps = 1e-6;
 39 LL MOD = 1000000007;
 40
 41 int n;
 42 double D, L;
 43 struct Point {
 44     double x, y;
 45     Point(double _x = 0, double _y = 0) {
 46         x = _x; y = _y;
 47     }
 48 };
 49
 50 double cross(Point a, Point b) {
 51     return a.x * b.y - a.y * b.x;
 52 }
 53
 54 struct Polygon {
 55     Point p[MAXN];
 56     int cnt;
 57     double get_area() {
 58         int pre = cnt;
 59         double ans = 0;
 60         rep (i, 1, cnt) {
 61             ans += cross(p[pre], p[i]);
 62             pre = i;
 63         }
 64         return fabs(ans / 2);
 65     }
 66
 67 }pol, po;
 68
 69 Point get_point(Point a, Point b, double y0) {
 70     if(fabs(a.x - b.x) < eps) return Point(a.x, y0);
 71     double bi = (y0 - a.y) / (b.y - a.y);
 72     return Point(a.x + bi * (b.x - a.x), a.y + bi * (b.y - a.y));
 73 }
 74
 75 double find_area(double y0) {
 76     int cnt = 0, pre = po.cnt;
 77     rep (i, 1, po.cnt) {
 78         if((y0 - po.p[pre].y) * (y0 - po.p[i].y) <= 0) {
 79             pol.p[++cnt] = get_point(po.p[pre], po.p[i], y0);
 80         }
 81         if(po.p[i].y < y0) pol.p[++cnt] = po.p[i];
 82         pre = i;
 83     }
 84     pol.cnt = cnt;
 85     return pol.get_area();
 86 }
 87
 88 int main()
 89 {
 90     //FIN;
 91     while(~scanf("%d", &po.cnt)) {
 92         cin >> D >> L;
 93         L *= 1000;
 94         double mx_y = 0.0;
 95         rep (i, 1, po.cnt) {
 96             scanf("%lf %lf", &po.p[i].x, &po.p[i].y);
 97             mx_y = max(mx_y, po.p[i].y);
 98         }
 99         double low = 0.0, high = mx_y;
100         while(high - low > eps) {
101             double mid = (low + high) / 2.0;
102             if(find_area(mid) * D < L) low = mid;
103             else high = mid;
104         }
105         printf("%.2lf\n", low);
106     }
107     return 0;
108 }
109
110 /**************************************************************
111     Problem: 1634
112     User: csust_Rush
113     Language: C++
114     Result: Accepted
115     Time:12 ms
116     Memory:4920 kb
117 ****************************************************************/

H：Restaurant Ratings

一个数位的简单DP，DP[i][j]表示i个人打j分的方案数

 1 #include <map>
 2 #include <set>
 3 #include <stack>
 4 #include <queue>
 5 #include <cmath>
 6 #include <ctime>
 7 #include <vector>
 8 #include <cstdio>
 9 #include <cctype>
10 #include <cstring>
11 #include <cstdlib>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 #define INF 0x3f3f3f3f
16 #define inf (-((LL)1<<40))
17 #define lson k<<1, L, (L + R)>>1
18 #define rson k<<1|1,  ((L + R)>>1) + 1, R
19 #define mem0(a) memset(a,0,sizeof(a))
20 #define mem1(a) memset(a,-1,sizeof(a))
21 #define mem(a, b) memset(a, b, sizeof(a))
22 #define FIN freopen("in.txt", "r", stdin)
23 #define FOUT freopen("out.txt", "w", stdout)
24 #define rep(i, a, b) for(int i = a; i <= b; i ++)
25 #define dec(i, a, b) for(int i = a; i >= b; i --)
26
27 template<class T> T CMP_MIN(T a, T b) { return a < b; }
28 template<class T> T CMP_MAX(T a, T b) { return a > b; }
29 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
30 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
33
34 //typedef __int64 LL;
35 typedef long long LL;
36 const int MAXN = 51000;
37 const int MAXM = 110000;
38 const double eps = 1e-4;
39 LL MOD = 1000000007;
40
41 LL dp[110][110];
42
43 void init_dp(int n) {
44     dp[0][0] = 1;
45     rep (i, 1, n) {
46         rep (j, 0, n) {
47             rep (k, 0, j) {
48                 dp[i][j] += dp[i - 1][j - k];
49             }
50         }
51     }
52 }
53
54 int n, a[110];
55
56 int main()
57 {
58     init_dp(100);
59     while(~scanf("%d", &n) && n) {
60         int sum = 0;
61         rep (i, 1, n) {
62             scanf("%d", &a[i]);
63             sum += a[i];
64         }
65         LL ans = 1;
66         rep (i, 0, sum - 1) {
67             ans += dp[n][i];
68         }
69         rep (i, 1, n) {
70             rep (j, 0, a[i] - 1) {
71                 ans += dp[n - i][sum - j];
72             }
73             sum -= a[i];
74         }
75         cout << ans << endl;
76     }
77     return 0;
78 }
79
80 /**************************************************************
81     Problem: 1635
82     User: csust_Rush
83     Language: C++
84     Result: Accepted
85     Time:16 ms
86     Memory:1580 kb
87 ****************************************************************/

I：Locked Treasure

据说是看样例找规律，然后发现ans=C(n, m-1)，不过也有类似的完美证明

J：Yet Satisfiability Again!

比赛时最后几分钟用暴力交了一发，发现AC了，赛后看其他人时间，大概都是暴利做的

最后干脆连题解都是用暴力啊= =

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <cstdlib>
 5 #include <vector>
 6 #include <algorithm>
 7 #include <cstring>
 8 #define mem0(a) memset(a, 0, sizeof(a))
 9 #define rep(i,a, b) for(int i = a; i <= b; i ++)
10 using namespace std;
11
12 int t, n, m;
13 struct Node {
14     int a[21], f[21], cnt;
15     bool judge(int s) {
16         int ans = 0, p = 0;
17         while(p < cnt && !ans) {
18             ans |= a[p] ^ (((1<<f[p]) & s) ? 1 : 0);
19             p ++;
20         }
21         return ans;
22     }
23 }mt[110];
24 char s[111100];
25
26 void handle(char *s, int id) {
27     int len = strlen(s), cnt = 0;
28     rep (i, 0, len - 1) {
29         if(s[i] == ‘X‘) {
30             int num = 0, p = i;
31             while(isdigit(s[++p])) num = num * 10 + s[p] - ‘0‘;
32             mt[id].a[cnt] = i > 0 && s[i - 1] == ‘~‘;
33             mt[id].f[cnt++] = num - 1;
34         }
35     }
36     mt[id].cnt = cnt;
37 }
38
39 int main()
40 {
41     //freopen("in.txt", "r", stdin);
42     while(cin >> t) while(t--) {
43         scanf("%d %d", &n, &m);
44         gets(s);
45         rep (i, 1, m) {
46             gets(s);
47             handle(s, i);
48         }
49         int ok = 0, h = (1 << n) - 1;
50         rep (s, 0, h) {
51             int p = 1;
52             rep (i, 1, m) {
53                 p &= mt[i].judge(s);
54                 if(!p) break;
55             }
56             ok |= p;
57             if(ok) {
58                 break;
59             }
60         }
61         if(ok) puts("satisfiable");
62         else puts("unsatisfiable");
63     }
64     return 0;
65 }
66
67 /**************************************************************
68     Problem: 1637
69     User: csust_Rush
70     Language: C++
71     Result: Accepted
72     Time:1768 ms
73     Memory:1612 kb
74 ****************************************************************/

K：Continued Fraction

做了两次了，第一次以为超LL，后来队友用BigInt（自己写的模板）过了

后来又听说LL可以过，再回头看自己代码，发现有用LL*LL，这必然超啊。。。改了后发现立马AC

 1 #include <stdio.h>
 2 #include <iostream>
 3 #include <cmath>
 4 #include <cstdlib>
 5 using namespace std;
 6
 7 typedef long long LL;
 8
 9 int n, m;
10 LL a1[100], a2[100];
11
12 LL gcd(LL a, LL b) {
13     return b == 0 ? a : gcd(b, a % b);
14 }
15
16 void dfs1(LL &a, LL &b, LL *arr, int i, int n)
17 {
18     if(i >= n - 1) {
19         a = 1; b = arr[i]; return ;
20     }
21     dfs1(a, b, arr, i + 1, n);
22     LL tmp = a;
23     a = b; b = arr[i] * b + tmp;
24     LL g = gcd(a, b);
25     a /= g; b /= g;
26 }
27
28 void add(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) {
29     x = A1 * B2 + A2 * B1;
30     y = B1 * B2;
31     LL g = gcd(x, y);
32     x /= g; y /= g;
33 }
34 void sub(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) {
35     x = A1 * B2 - A2 * B1;
36     y = B1 * B2;
37     LL g = gcd(x, y);
38     x /= g; y /= g;
39 }
40 void mul(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) {
41     x = A1 * A2; y = B1 * B2;
42     LL g = gcd(x, y);
43     x /= g; y /= g;
44 }
45 void div(LL A1, LL B1, LL A2, LL B2, LL &x, LL &y) {
46     x = A1 * B2; y = A2 * B1;
47     LL g = gcd(x, y);
48     x /= g; y /= g;
49 }
50
51 void print(LL x, LL y) {
52     if (x > 0 && y < 0 || x < 0 && y > 0) {
53         if (x % y != 0) cout<< x / y - 1 << " ";
54         else {
55             cout << x / y << endl;;
56             return;
57         }
58         x = abs(y) - abs(x) % abs(y);
59         print(abs(y), x);
60         return ;
61     }
62     cout << x / y;
63     if(x % y) {
64         printf(" ");
65         print(y, x % y);
66     }
67     else printf("\n");
68 }
69
70 int main()
71 {
72     //freopen("in.txt", "r", stdin);
73     int cas = 0;
74     while(~scanf("%d %d", &n, &m)) {
75         for(int i = 0; i < n; i ++) cin >> a1[i];
76         for(int i = 0; i < m; i ++) cin >> a2[i];
77         LL A1 = 0, B1 = 1, A2 = 0, B2 = 1;
78         if (n > 1) dfs1(A1, B1, a1, 1, n);
79         if (m > 1) dfs1(A2, B2, a2, 1, m);
80         A1 = a1[0] * B1 + A1;
81         A2 = a2[0] * B2 + A2;
82         LL x, y;
83         //printf("Case %d:\n", ++cas);
84         add(A1, B1, A2, B2, x, y); print(x, y);
85         sub(A1, B1, A2, B2, x, y); print(x, y);
86         mul(A1, B1, A2, B2, x, y); print(x, y);
87         div(A1, B1, A2, B2, x, y); print(x, y);
88     }
89     return 0;
90 }
91

时间： 2024-10-08 00:09:07

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