网络流24题5

Posted 2020-10-27 tjucxz

https://loj.ac/problem/6004

二分图多重匹配

#include <iostream>
#include <stdio.h>
#include <queue>
#include <cstring>
using namespace std;
const int maxn = 1000;
int head[maxn];
int tot = 0;
struct edge{
    int v,nex,w;
}e[maxn*120];
void addedge(int u,int v,int w){
    e[tot] = (edge){v,head[u],w};
    head[u] = tot++;
    e[tot] = (edge){u,head[v],0};
    head[v] = tot++;
}
int deep[maxn];
bool bfs(int S,int T){
    memset(deep,0,sizeof(deep));
    deep[S] = 1;
    queue<int> q;
    q.push(S);
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i=head[now];i!=-1;i=e[i].nex){
            int v = e[i].v;
            int w = e[i].w;
            if(deep[v]!=0 || w<=0) continue;
            deep[v] = deep[now]+1;
            q.push(v);
        }
    }
    return deep[T];
}
int dfs(int now,int T,int maxflow){
    if(now==T) return maxflow;
    int all = 0;
    for(int i=head[now];i!=-1 && all<maxflow;i=e[i].nex){
        int v = e[i].v;
        int w = e[i].w;
        if(deep[v]!=deep[now]+1 || w<=0) continue;
        int tt = dfs(v,T,min(maxflow-all,w));
        e[i].w-=tt;
        e[i^1].w+=tt;
        all+=tt;
    }
    return all;
}

int dinic(int S,int T){
    int ret = 0;
    while(bfs(S,T)){
        ret+=dfs(S,T,0x3f3f3f3f);
    }
    return ret;
}
int main()
{
    int m,n;
    scanf("%d%d",&m,&n);
    memset(head,-1,sizeof(head));
    int all = 0;
    for(int i=1;i<=m;i++){
        int t ;
        scanf("%d",&t);
        all+=t;
        addedge(0,i,t);
    }
    for(int i=1;i<=n;i++){
        int t;
        scanf("%d",&t);
        addedge(m+i,m+n+1,t);
    }
    for(int i=1;i<=m;i++){
        for(int j=m+1;j<=m+n;j++){
            addedge(i,j,1);
        }
    }
    int ans = dinic(0,m+n+1);
    if(all==ans){
        printf("1\n");
        for(int i=1;i<=m;i++){
            for(int j=head[i];j!=-1;j=e[j].nex){
                int v = e[j].v;
                int w = e[j].w;
                if(v<=m+n && v>=m+1 && w==0) {
                    printf("%d ",v-m);
                }
            }
            printf("\n");
        }

    }else{
        printf("0\n");
    }
    return 0;
}