HDU 4289 Control（网络流）

题意：有N个城市 M条无向边　有一恐怖分子要从某一城市到另一城市　打算在某些城市安放一些SA 去抓住他但若在某个城市安放SA需要一定费用 求要抓到恐怖分子 最少的费用是多少？

思路：网络流问题。建一超级源点和汇点与原源点、汇点相连，然后把一个城市拆成两个点 边权为其费用两相连城市间的边权为无穷大  求其最大流即可。

//62MS   1176K

#include <stdio.h>
#include <string.h>
#define L(u) ((u) << 1)
#define R(u) ((u) << 1 | 1)
#define VM 420
#define EM 100000
#define inf 0x3f3f3f3f
struct E
{
    intto,cap,nxt;
}edge[EM];

int head[VM],gap[VM],dist[VM],cur[VM],pre[VM];
int e,src,des,n,m;
void addedge (int cu,int cv,int cw)
{

    edge[e].to =cv;
    edge[e].cap= cw;
    edge[e].nxt= head[cu];
    head[cu] =e;
    e ++;
    edge[e].to =cu;
    edge[e].cap= 0;
    edge[e].nxt= head[cv];
    head[cv] =e;
    e ++;
}
int min (int a ,int b)
{
   return a > b ?b : a;
}

int sap ()
{
    memset(dist,0,sizeof(dist));
    memset(gap,0,sizeof (dist));
    memcpy(cur,head,sizeof(dist));
    int res =0;
    int u =pre[src] = src;
    int aug =inf;
    gap[0] =n;
    while(dist[src] < n)
    {
loop:
       for (int &i = cur[u];i != -1;i = edge[i].nxt)
       {
           int v = edge[i].to;
           if (edge[i].cap && dist[u] ==dist[v] + 1)
           {
               aug = min (aug,edge[i].cap);
               pre[v] = u;
               u = v;
               if (v == des)
               {
                   res += aug;
                   for (u = pre[u];v != src;v = u,u = pre[u])
                   {
                       edge[cur[u]].cap -= aug;
                       edge[cur[u]^1].cap += aug;
                   }
                   aug = inf; //
               }
               goto loop;
           }
       }
       int mindist = n;  //
       for (int i = head[u];i != -1;i = edge[i].nxt)
       {
           int v = edge[i].to;
           if (edge[i].cap && mindist> dist[v])
           {
               cur[u] = i;
               mindist = dist[v];
           }
       }
       if ((--gap[dist[u]]) == 0)
           break;
       dist[u] = mindist + 1;
       gap[dist[u]] ++;
       u = pre[u];
    }
    returnres;
}

int main()
{
    int i;
    while(~scanf("%d%d", &n, &m))
    {
       e = 0;
       memset(head,-1, sizeof(head));
      scanf("%d%d", &src, &des);
       addedge(0,L(src), inf);
      addedge(R(des), R(n + 1), inf);
       src =0;
       des = R(n +1);
       int u, v,c;

       for (i=1;i<=n; ++i)
       {
          scanf("%d",&c);
         addedge(L(i), R(i), c);
         addedge(R(i), L(i), c);
       }
       for (i=1;i<=m; ++i)
       {
         scanf("%d%d", &u, &v);
         addedge(R(u), L(v), inf);
         addedge(R(v), L(u), inf);
       }
       n =2*(n+1);
      printf("%d\n", sap());
    }
    return0;
}