hdu 2686最小费用最大流问题

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
#define inf 0x7fffffff
struct node{
    int u,v,w,f,next;
}bian[20000];
int dis[6000],pre[6000],head[6000],visit[6000],yong,minf;
void build(int u,int v,int w,int f) {
    bian[yong].u=u;
    bian[yong].v=v;
    bian[yong].w=w;
    bian[yong].f=f;
    bian[yong].next=head[u];
    head[u]=yong++;
}
void adde(int u,int v,int w,int f) {
    build(u,v,w,f);
    build(v,u,-w,0);
}
int spfa(int s,int t) {
    memset(visit,0,sizeof(visit));
    memset(pre,-1,sizeof(pre));
    for(int i=0;i<=t;i++)//注意是从0开始而不是从s开始
        dis[i]=inf;
    queue<int>q;
     visit[s]=1;//必须是s
     q.push(s);
     dis[s]=0;
     while(!q.empty ()) {
         int u=q.front ();
         for(int index=head[u];index!=-1;index=bian[index].next) {
             int v=bian[index].v;
             if(bian[index].f&&dis[v]>dis[u]+bian[index].w) {
                 dis[v]=dis[u]+bian[index].w;
                 pre[v]=index;
                 if(!visit[v]) {
                     visit[v]=1;
                     q.push (v);
                 }
             }
         }
         q.pop();
         visit[u]=0;//是u
     }
     if(dis[t]==inf)
         return 0;
     return 1;
}
void addf(int s,int t ) {
    int i=pre[t];
    int j;
    while(i!=-1) {
        j=i^1;
        bian[i].f--;
        bian[j].f++;
        minf+=bian[i].w;
        i=pre[bian[i].u];//前一个u
    }
}
int main() {
    int n,k,map[60][60];
    while(scanf("%d",&n)!=EOF) {
        memset(head,-1,sizeof(head));
        k=2;
        yong=0;
        int s=2*n*n;
        int t=s+1;
        int i,j;
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                scanf("%d",&map[i][j]);
            for(i=0;i<n;i++)
                for(j=0;j<n;j++) {
                    int b=i*n+j;
                    adde(2*b,2*b+1,-map[i][j],1);
                    adde(2*b,2*b+1,0,k-1);
                }
                for(i=0;i<n;i++)
                    for(j=0;j<n-1;j++) {
                      int b=i*n+j;
                      adde(2*b+1,2*(b+1),0,k);
                    }
                    for(i=0;i<n-1;i++)
                        for(j=0;j<n;j++) {
                            int b=i*n+j;
                            adde(2*b+1,2*(b+n),0,k);
                        }
                            adde(s,0,0,k);
                            adde(2*n*n-1,t,0,k);
                            minf=0;
                            while(spfa(s,t)) {
                                addf(s,t);
                            }
                            printf("%d\n",-minf);
                        }
                        return 0;
    }