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  • 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
    ",-minf);
                            }
                            return 0;
        }
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  • 原文地址:https://www.cnblogs.com/thefirstfeeling/p/4410993.html
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