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。
  1 #include<iostream>
  2 #include<vector>
  3 #include<queue>
  4 #include<stack>
  5 #include<cstdio>
  6 #include<cstring>
  7 using namespace std;
  8 const int maxn=1e3+5,INF=1e9;
  9 struct Edge{
 10     int from,to,cap,flow;
 11     Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
 12 };
 13 struct ISAP{
 14     int n,s,t;
 15     vector<Edge> edges;
 16     vector<int> G[maxn];
 17     bool vis[maxn];
 18     int d[maxn];
 19     int cur[maxn];
 20     int p[maxn];
 21     int num[maxn];
 22 
 23     void init(int n){
 24         this->n=n;
 25         edges.clear();
 26         for(int i=0;i<n;i++)G[i].clear();
 27     }
 28 
 29     void AddEdge(int from,int to,int cap){
 30         edges.push_back(Edge(from,to,cap,0));
 31         edges.push_back(Edge(to,from,0,0));
 32         G[from].push_back(edges.size()-2);
 33         G[to].push_back(edges.size()-1);
 34     }
 35 
 36     void BFS(){
 37         memset(vis,0,sizeof(vis));
 38         queue<int> Q;
 39         Q.push(t);
 40         d[t]=0;
 41         vis[t]=1;
 42         while(!Q.empty()){
 43             int x=Q.front();Q.pop();
 44             for(int i=0;i<G[x].size();i++){
 45                 Edge &e=edges[G[x][i]];
 46                 if(!vis[e.to]){
 47                     vis[e.to]=1;
 48                     d[e.to]=d[x]+1;
 49                     Q.push(e.to);
 50                 }
 51             }
 52         }
 53     }
 54 
 55     int Augment(){
 56         int x=t,a=INF;
 57         while(x!=s){
 58             Edge &e=edges[p[x]];
 59             a=min(a,e.cap-e.flow);
 60             x=edges[p[x]].from;
 61         }
 62         x=t;
 63         while(x!=s){
 64             edges[p[x]].flow+=a;
 65             edges[p[x]^1].flow-=a;
 66             x=edges[p[x]].from;
 67         }
 68         return a;
 69     }
 70 
 71     int Maxflow(int s,int t){
 72         this->s=s;this->t=t;
 73         int flow=0;
 74         BFS();
 75         memset(num,0,sizeof(num));
 76         for(int i=0;i<n;i++)num[d[i]]++;
 77         int x=s;
 78         memset(cur,0,sizeof(cur));
 79         while(d[s]<n){
 80             if(x==t){
 81                 flow+=Augment();
 82                 x=s;
 83             }
 84             int ok=0;
 85             for(int i=cur[x];i<G[x].size();i++){
 86                 Edge &e=edges[G[x][i]];
 87                 if(e.cap>e.flow&&d[x]==d[e.to]+1){
 88                     ok=1;
 89                     p[e.to]=G[x][i];
 90                     cur[x]=i;
 91                     x=e.to;
 92                     break;
 93                 }
 94             }
 95             if(!ok){
 96                 int m=n-1;
 97                 for(int i=0;i<G[x].size();i++){
 98                     Edge &e=edges[G[x][i]];
 99                     if(e.cap>e.flow)m=min(m,d[e.to]);
100                 }
101                 if(--num[x]==0)break;
102                 num[d[x]=m+1]++;
103                 cur[x]=0;
104                 if(x!=s)x=edges[p[x]].from;
105             }
106         }
107         return flow;
108     }
109 }isap;
110 
111 int main(){
112 //    freopen("input.txt","r",stdin);
113     //freopen("output.txt","w",stdout);
114     int n,m,s,t;
115     scanf("%d%d",&m,&n);
116     isap.init(n+1);
117     for(int i=1;i<=m;i++)isap.AddEdge(0,i,1);
118     for(int i=m+1;i<=n;i++)isap.AddEdge(i,n+1,1);
119     while(1){
120         scanf("%d%d",&s,&t);
121         if(s==-1&&t==-1)break;
122         isap.AddEdge(s,t,1);
123     }
124     int M=isap.Maxflow(0,n+1);
125     if(M==0){
126         printf("No Solution!
");
127     }else{
128         printf("%d
",M);
129         for(int i=0;i<isap.edges.size();i++){
130             if(isap.edges[i].from!=0&&isap.edges[i].to!=n+1&&isap.edges[i].flow==1){
131                 printf("%d %d
",isap.edges[i].from,isap.edges[i].to);
132             }
133         }
134     }
135     return 0;
136 }