BZOJ3438 小M的作物

叫什么"最大权闭合图" -- from PoPoQQQ

首先建立源点汇点S、T表示种在a，b两片土里，流量为收益。

然后每次读到一个集合的时候，新建两个虚拟节点si、ti，表示都种在a，b里的收益

S向si连边，ti向T连边，流量为收益，然后si、ti向集合内所有点连边。

之后跑一边最大流，ans = 总收益 - 最小割

 

  1 /**************************************************************
  2     Problem: 3438
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:2664 ms
  7     Memory:24280 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13  
 14 using namespace std;
 15 const int inf = (int) 1e9;
 16 const int N = 3005;
 17 const int M = 2000005;
 18  
 19 struct edges{
 20     int next, to, f;
 21     edges() {}
 22     edges(int _n, int _t, int _f) : next(_n), to(_t), f(_f) {}
 23 } e[M];
 24  
 25 int n, m;
 26 int cnt, ans, S, T;
 27 int first[N], tot = 1;
 28 int q[N], d[N];
 29  
 30 inline int read() {
 31     int x = 0;
 32     char ch = getchar();
 33     while (ch < '0' || '9' < ch)
 34         ch = getchar();
 35     while ('0' <= ch && ch <= '9') {
 36         x = x * 10 + ch - '0';
 37         ch = getchar();
 38     }
 39     return x;
 40 }
 41  
 42 inline void Add_Edges(int x, int y, int z){
 43     e[++tot] = edges(first[x], y, z), first[x] = tot;
 44     e[++tot] = edges(first[y], x, 0), first[y] = tot;
 45 }
 46    
 47 bool bfs(){
 48     memset(d, 0, sizeof(d));
 49     q[1] = S, d[S] = 1;
 50     int l = 0, r = 1, x, y;
 51     while (l < r){
 52         ++l;
 53         for (x = first[q[l]]; x; x = e[x].next){
 54             y = e[x].to;
 55             if (!d[y] && e[x].f)
 56                 q[++r] = y, d[y] = d[q[l]] + 1;
 57         }
 58     }
 59     return d[T];
 60 }
 61    
 62 int dinic(int p, int limit){
 63     if (p == T || !limit) return limit;
 64     int x, y, tmp, rest = limit;
 65     for (x = first[p]; x; x = e[x].next){
 66         y = e[x].to;
 67         if (d[y] == d[p] + 1 && e[x].f && rest){
 68             tmp = dinic(y, min(rest, e[x].f));
 69             rest -= tmp;
 70             e[x].f -= tmp, e[x ^ 1].f += tmp;
 71             if (!rest) return limit;
 72         }
 73     }
 74     if (limit == rest) d[p] = 0;
 75     return limit - rest;
 76 }
 77    
 78 int Dinic(){
 79     int res = 0, x;
 80     while (bfs())
 81         res += dinic(S, inf);
 82     return res;
 83 }
 84  
 85 int main() {
 86     int i, j, x, y, z;
 87     n = cnt = read(), S = 3001, T = 3002;
 88     for (i = 1; i <= n; ++i) {
 89         ans += (x = read());
 90         Add_Edges(S, i, x);
 91     }
 92     for (i = 1; i <= n; ++i) {
 93         ans += (x = read());
 94         Add_Edges(i, T, x);
 95     }
 96     m = read();
 97     for (i = 1; i <= m; ++i) {
 98         j = read(), x = read(), y = read();
 99         ans += x + y;
100         Add_Edges(S, ++cnt, x), Add_Edges(++cnt, T, y);
101         while (j--) {
102             z = read();
103             Add_Edges(z, cnt, inf), Add_Edges(cnt - 1, z, inf);
104         }
105     }
106     printf("%d\n", ans - Dinic());
107     return 0;
108 }
View Code

 

posted on 2014-11-30 18:04  Xs酱~  阅读(561)  评论(0)  编辑  收藏  举报