hdu 4289 Control (成都网络赛最小割)

    题目： http://acm.hdu.edu.cn/showproblem.php?pid=4289

   

   分析：题目给定的是每个节点的权值，所以要进行拆点操作，即在a和a+n之间连边，流量为点权。然后对于原来的边的情况，建立四条边（无向），（a+n，b）和（b+n，a），流量为无穷，以及它们各自的反向边（b，a+n）和（a，b+n），流量为0，最后把汇点改为输入值+n。

   代码：

#include <stdio.h>#include <string.h>#define N 10010#define M 400010const int inf = 0x3f3f3f3f;struct E{    int to, frm, nxt, cap;}edge[M];int head[N],e,n,m,src,des;int dep[N], gap[N];void addedge(int u, int v, int w){    edge[e].frm = u;    edge[e].to = v;    edge[e].cap = w;    edge[e].nxt = head[u];    head[u] = e++;    edge[e].frm = v;    edge[e].to = u;    edge[e].cap = 0;    edge[e].nxt = head[v];    head[v] = e++;}int que[N];void BFS(){    memset(dep, -1, sizeof(dep));    memset(gap, 0, sizeof(gap));    gap[0] = 1;    int front = 0, rear = 0;    dep[des] = 0;    que[rear++] = des;    int u, v;    while (front != rear)    {        u = que[front++];        front = front%N;        for (int i=head[u]; i!=-1; i=edge[i].nxt)        {            v = edge[i].to;            if (edge[i].cap != 0 || dep[v] != -1)                continue;            que[rear++] = v;            rear = rear % N;            ++gap[dep[v] = dep[u] + 1];        }    }}int cur[N],stack[N];int Sap()      {    int res = 0;    BFS();    int top = 0;    memcpy(cur, head, sizeof(head));    int u = src, i;    while (dep[src] < n*2)    {        if (u == des)        {            int temp = inf, inser = n;            for (i=0; i!=top; ++i)                if (temp > edge[stack[i]].cap)                {                    temp = edge[stack[i]].cap;                    inser = i;                }            for (i=0; i!=top; ++i)            {                edge[stack[i]].cap -= temp;                edge[stack[i]^1].cap += temp;            }            res += temp;            top = inser;            u = edge[stack[top]].frm;        }        if (u != des && gap[dep[u] -1] == 0)            break;        for (i = cur[u]; i != -1; i = edge[i].nxt)            if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1)                break;        if (i != -1)        {            cur[u] = i;            stack[top++] = i;            u = edge[i].to;        }        else        {            int min = n*2;            for (i = head[u]; i != -1; i = edge[i].nxt)            {                if (edge[i].cap == 0)                    continue;                if (min > dep[edge[i].to])                {                    min = dep[edge[i].to];                    cur[u] = i;                }            }            --gap[dep[u]];            ++gap[dep[u] = min + 1];            if (u != src)                u = edge[stack[--top]].frm;        }    }    return res;}int main(){    int i,c,x,y;while(scanf("%d%d",&n,&m)!=EOF){   e=0;memset(head,-1,sizeof(head));    scanf("%d%d",&src,&des);des+=n;        for(i=1;i<=n;i++){   scanf("%d",&c);   addedge(i,i+n,c);}for(i=1;i<=m;i++){   scanf("%d%d",&x,&y);   addedge(x+n,y,inf);   addedge(y+n,x,inf);}printf("%d\n",Sap());}return 0;}