hdu 4289 Control 网络流

题目链接

给出一些点， 每个点有一个权值， 给出一些边， 起点以及终点， 去掉一些点使得起点和终点不连通， 求最小的val。

拆点， 把一个点s拆成s和s'， 之间建一条边， 权值为点权。 对于一条边， <u, v> 建边<u, v'>, <v, u'> 权值为inf， 跑一遍最大流。

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define pb(x) push_back(x)
  4 #define ll long long
  5 #define mk(x, y) make_pair(x, y)
  6 #define lson l, m, rt<<1
  7 #define mem(a) memset(a, 0, sizeof(a))
  8 #define rson m+1, r, rt<<1|1
  9 #define mem1(a) memset(a, -1, sizeof(a))
 10 #define mem2(a) memset(a, 0x3f, sizeof(a))
 11 #define rep(i, a, n) for(int i = a; i<n; i++)
 12 #define ull unsigned long long
 13 typedef pair<int, int> pll;
 14 const double PI = acos(-1.0);
 15 const double eps = 1e-8;
 16 const int mod = 1e9+7;
 17 const int inf = 1061109567;
 18 const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
 19 const int maxx = 405;
 20 const int maxn = 40005;
 21 int head[maxn*4], s, t, num, q[maxn*3], dis[maxn];
 22 struct node
 23 {
 24     int to, nextt, c;
 25 }e[maxn*4];
 26 void init() {
 27     mem1(head);
 28     num = 0;
 29 }
 30 void add(int u, int v, int c) {
 31     e[num].to = v; e[num].nextt = head[u]; e[num].c = c; head[u] = num++;
 32     e[num].to = u; e[num].nextt = head[v]; e[num].c = 0; head[v] = num++;
 33 }
 34 int bfs() {
 35     int u, v, st = 0, ed = 0;
 36     mem(dis);
 37     dis[s] = 1;
 38     q[ed++] = s;
 39     while(st<ed) {
 40         u = q[st++];
 41         for(int i = head[u]; ~i; i = e[i].nextt) {
 42             v = e[i].to;
 43             if(e[i].c&&!dis[v]) {
 44                 dis[v] = dis[u]+1;
 45                 if(v == t)
 46                     return 1;
 47                 q[ed++] = v;
 48             }
 49         }
 50     }
 51     return 0;
 52 }
 53 int dfs(int u, int limit) {
 54     if(u == t)
 55         return limit;
 56     int cost = 0;
 57     for(int i = head[u]; ~i; i = e[i].nextt) {
 58         int v = e[i].to;
 59         if(e[i].c&&dis[u] == dis[v]-1) {
 60             int tmp = dfs(v, min(limit-cost, e[i].c));
 61             if(tmp>0) {
 62                 e[i].c -= tmp;
 63                 e[i^1].c += tmp;
 64                 cost += tmp;
 65                 if(cost == limit)
 66                     break;
 67             } else {
 68                 dis[v] = -1;
 69             }
 70         }
 71     }
 72     return cost;
 73 }
 74 int dinic() {
 75     int ans = 0;
 76     while(bfs()) {
 77         ans += dfs(s, inf);
 78     }
 79     return ans;
 80 }
 81 int main()
 82 {
 83     int n, m, val, x, y;
 84     while(cin>>n>>m) {
 85         cin>>s>>t;
 86         t+=n;
 87         init();
 88         for(int i = 1; i<=n; i++) {
 89             scanf("%d", &val);
 90             add(i, i+n, val);
 91         }
 92         while(m--) {
 93             scanf("%d%d", &x, &y);
 94             add(x+n, y, inf);
 95             add(y+n, x, inf);
 96         }
 97         int ans = dinic();
 98         cout<<ans<<endl;
 99     }
100 }

 

posted on 2015-12-03 23:10  yohaha  阅读(364)  评论(0)  编辑  收藏  举报

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