网络流24题——飞行员配对方案问题(二分图最大匹配)

链接：https://www.oj.swust.edu.cn/oj/problem/show/1736

分析：模板题，增加源点和汇点s、t，从s向U中所有点连一条容量为1的边，从V中所有点向t中所有点连一条容量为1的边，然后求s-t最大流即为所求。最大流用的isap模板。

#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1e3+5,INF=1e9;
struct Edge{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct ISAP{
    int n,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];
    int p[maxn];
    int num[maxn];

    void init(int n){
        this->n=n;
        edges.clear();
        for(int i=0;i<n;i++)G[i].clear();
    }

    void AddEdge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));
        G[from].push_back(edges.size()-2);
        G[to].push_back(edges.size()-1);
    }

    void BFS(){
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        Q.push(t);
        d[t]=0;
        vis[t]=1;
        while(!Q.empty()){
            int x=Q.front();Q.pop();
            for(int i=0;i<G[x].size();i++){
                Edge &e=edges[G[x][i]];
                if(!vis[e.to]){
                    vis[e.to]=1;
                    d[e.to]=d[x]+1;
                    Q.push(e.to);
                }
            }
        }
    }

    int Augment(){
        int x=t,a=INF;
        while(x!=s){
            Edge &e=edges[p[x]];
            a=min(a,e.cap-e.flow);
            x=edges[p[x]].from;
        }
        x=t;
        while(x!=s){
            edges[p[x]].flow+=a;
            edges[p[x]^1].flow-=a;
            x=edges[p[x]].from;
        }
        return a;
    }

    int Maxflow(int s,int t){
        this->s=s;this->t=t;
        int flow=0;
        BFS();
        memset(num,0,sizeof(num));
        for(int i=0;i<n;i++)num[d[i]]++;
        int x=s;
        memset(cur,0,sizeof(cur));
        while(d[s]<n){
            if(x==t){
                flow+=Augment();
                x=s;
            }
            int ok=0;
            for(int i=cur[x];i<G[x].size();i++){
                Edge &e=edges[G[x][i]];
                if(e.cap>e.flow&&d[x]==d[e.to]+1){
                    ok=1;
                    p[e.to]=G[x][i];
                    cur[x]=i;
                    x=e.to;
                    break;
                }
            }
            if(!ok){
                int m=n-1;
                for(int i=0;i<G[x].size();i++){
                    Edge &e=edges[G[x][i]];
                    if(e.cap>e.flow)m=min(m,d[e.to]);
                }
                if(--num[x]==0)break;
                num[d[x]=m+1]++;
                cur[x]=0;
                if(x!=s)x=edges[p[x]].from;
            }
        }
        return flow;
    }
}isap;

int main(){
//    freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int n,m,s,t;
    scanf("%d%d",&m,&n);
    isap.init(n+1);
    for(int i=1;i<=m;i++)isap.AddEdge(0,i,1);
    for(int i=m+1;i<=n;i++)isap.AddEdge(i,n+1,1);
    while(1){
        scanf("%d%d",&s,&t);
        if(s==-1&&t==-1)break;
        isap.AddEdge(s,t,1);
    }
    int M=isap.Maxflow(0,n+1);
    if(M==0){
        printf("No Solution!
");
    }else{
        printf("%d
",M);
        for(int i=0;i<isap.edges.size();i++){
            if(isap.edges[i].from!=0&&isap.edges[i].to!=n+1&&isap.edges[i].flow==1){
                printf("%d %d
",isap.edges[i].from,isap.edges[i].to);
            }
        }
    }
    return 0;
}