• 网络流24题——试题库问题(二分图多重匹配)


    链接:https://www.oj.swust.edu.cn/oj/problem/show/1742

    分析:裸题,直接上模板搞一下二分图匹配即可。

      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 int k,n,m=0;
     10 struct Edge{
     11     int from,to,cap,flow;
     12     Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
     13 };
     14 struct ISAP{
     15     int n,s,t;
     16     vector<Edge> edges;
     17     vector<int> G[maxn];
     18     bool vis[maxn];
     19     int d[maxn];
     20     int cur[maxn];
     21     int p[maxn];
     22     int num[maxn];
     23 
     24     void init(int n){
     25         this->n=n;
     26         edges.clear();
     27         for(int i=0;i<n;i++)G[i].clear();
     28     }
     29 
     30     void AddEdge(int from,int to,int cap){
     31         edges.push_back(Edge(from,to,cap,0));
     32         edges.push_back(Edge(to,from,0,0));
     33         G[from].push_back(edges.size()-2);
     34         G[to].push_back(edges.size()-1);
     35     }
     36 
     37     void BFS(){
     38         memset(vis,0,sizeof(vis));
     39         queue<int> Q;
     40         Q.push(t);
     41         d[t]=0;
     42         vis[t]=1;
     43         while(!Q.empty()){
     44             int x=Q.front();Q.pop();
     45             for(int i=0;i<G[x].size();i++){
     46                 Edge &e=edges[G[x][i]^1];
     47                 if(e.flow==e.cap)continue;
     48                 if(!vis[e.from]){
     49                     vis[e.from]=1;
     50                     d[e.from]=d[x]+1;
     51                     Q.push(e.from);
     52                 }
     53             }
     54         }
     55     }
     56 
     57     int Augment(){
     58         int x=t,a=INF;
     59         while(x!=s){
     60             Edge &e=edges[p[x]];
     61             a=min(a,e.cap-e.flow);
     62             x=edges[p[x]].from;
     63         }
     64         x=t;
     65         while(x!=s){
     66             edges[p[x]].flow+=a;
     67             edges[p[x]^1].flow-=a;
     68             x=edges[p[x]].from;
     69         }
     70         return a;
     71     }
     72 
     73     int Maxflow(int s,int t){
     74         this->s=s;this->t=t;
     75         int flow=0;
     76         BFS();
     77         memset(num,0,sizeof(num));
     78         for(int i=0;i<n;i++)num[d[i]]++;
     79         int x=s;
     80         memset(cur,0,sizeof(cur));
     81         while(d[s]<n){
     82             if(x==t){
     83                 flow+=Augment();
     84                 x=s;
     85             }
     86             int ok=0;
     87             for(int i=cur[x];i<G[x].size();i++){
     88                 Edge &e=edges[G[x][i]];
     89                 if(e.cap>e.flow&&d[x]==d[e.to]+1){
     90                     ok=1;
     91                     p[e.to]=G[x][i];
     92                     cur[x]=i;
     93                     x=e.to;
     94                     break;
     95                 }
     96             }
     97             if(!ok){
     98                 int m=n-1;
     99                 for(int i=0;i<G[x].size();i++){
    100                     Edge &e=edges[G[x][i]];
    101                     if(e.cap>e.flow)m=min(m,d[e.to]);
    102                 }
    103                 if(--num[d[x]]==0)break;
    104                 num[d[x]=m+1]++;
    105                 cur[x]=0;
    106                 if(x!=s)x=edges[p[x]].from;
    107             }
    108         }
    109         return flow;
    110     }
    111     void Print(){
    112         for(int i=1;i<=k;i++){
    113             printf("%d:",i);
    114             for(int j=0;j<G[i].size();j++){
    115                 Edge &e=edges[G[i][j]];
    116                 if(e.flow==1){
    117                     printf(" %d",e.to-k);
    118                 }
    119             }
    120             printf("
    ");
    121         }
    122     }
    123 }isap;
    124 int main(){
    125 //    freopen("e:\in.txt","r",stdin);
    126     int v,p;
    127     scanf("%d%d",&k,&n);
    128     isap.init(k+n+2);
    129     for(int i=1;i<=k;i++){
    130         scanf("%d",&v);
    131         isap.AddEdge(0,i,v);
    132         m+=v;
    133     }
    134     for(int i=k+1;i<=k+n;i++){
    135         scanf("%d",&p);
    136         isap.AddEdge(i,k+n+1,1);
    137         while(p--){
    138             scanf("%d",&v);
    139             isap.AddEdge(v,i,1);
    140         }
    141     }
    142     if(isap.Maxflow(0,k+n+1)<m){
    143         printf("No Solution!
    ");
    144     }else{
    145         isap.Print();
    146     }
    147 
    148     return 0;
    149 }
  • 相关阅读:
    绑定源特性的行为
    理解IOC和DI
    相遇~~~~
    选美素数~~~~~~~(⊙o⊙)
    狐狸找兔纸
    多重背包Q
    最长升序子序列
    构造逆序数
    田径赛马
    01背包M
  • 原文地址:https://www.cnblogs.com/7391-KID/p/7448130.html
Copyright © 2020-2023  润新知