hdu3605 Escape 二分图多重匹配/最大流

2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine what all of people can live in these planets.

题意：有N个人，每个人能适应若干个星球（共10个星球），每个星球能够容纳若干人，问是否能让所有人都有星球住。

其实应该是一个二分图多重匹配的题目。

不过一开始当成最大流的题目了，做的时候TLE了，然后将对10个星球的适应性的情况用状压变成了2^10种情况，即2^10种人，并且加上读入优化在vj上也卡过去了，不过在杭电上始终还是TLE了。

 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #include<algorithm>
 using namespace std;

 const int maxn=;
 const int maxm=;
 int match[maxn][maxm],g[maxn][maxm],cnt[maxm],lim[maxn],n,m;
 bool vis[maxm];

 bool dfs(int s){
     for(int i=;i<m;++i){
         if(g[s][i]==||vis[i]==)continue;
         vis[i]=;
         if(cnt[i]<lim[i]){
             match[i][cnt[i]++]=s;
             return ;
         }
         else{
             for(int j=;j<lim[i];++j){
                 if(dfs(match[i][j])==){
                     match[i][j]=s;
                     return ;
                 }
             }
         }
     }
     return ;
 }
 bool hungary(int n){
     for(int i=;i<n;++i){
         memset(vis,,sizeof(vis));
         if(dfs(i)==)return ;
     }
     return ;
 }
 int main(){
     while(scanf("%d%d",&n,&m)!=EOF){
         memset(cnt,,sizeof(cnt));
         for(int i=;i<n;++i)
             for(int j=;j<m;++j)scanf("%d",&g[i][j]);
         for(int i=;i<m;++i)scanf("%d",&lim[i]);
         if(hungary(n))printf("YES\n");
         else printf("NO\n");
     }
     return ;
 }

多重匹配

 #pragma comment(linker,"/STACK:16777216")
 #include<stdio.h>
 #include<string.h>
 #include<vector>
 #include<queue>
 #include<algorithm>
 using namespace std;
 const int maxm=;
 const int INF=0x3f3f3f3f;

 struct edge{
     int from,to,c,f;
     edge(int a,int b,int m,int n):from(a),to(b),c(m),f(n){}
 };

 struct dinic{
     int s,t,m;
     vector<edge>e;
     vector<int>g[maxm];
     bool vis[maxm];
     int cur[maxm],d[maxm];

     void init(int n){
         for(int i=;i<=n;i++)g[i].clear();
         e.clear();
     }

     void add(int a,int b,int c){
         e.push_back(edge(a,b,c,));
         e.push_back(edge(b,a,,));
         m=e.size();
         g[a].push_back(m-);
         g[b].push_back(m-);
     }

     bool bfs(){
         memset(vis,,sizeof(vis));
         queue<int>q;
         q.push(s);
         vis[s]=;
         d[s]=;
         while(!q.empty()){
             int u=q.front();
             q.pop();
             for(int i=;i<g[u].size();i++){
                 edge tmp=e[g[u][i]];
                 if(!vis[tmp.to]&&tmp.c>tmp.f){
                     d[tmp.to]=d[u]+;
                     vis[tmp.to]=;
                     q.push(tmp.to);
                 }
             }
         }
         return vis[t];
     }

     int dfs(int x,int a){
         if(x==t||a==)return a;
         int flow=,f;
         for(int& i=cur[x];i<g[x].size();i++){
             edge& tmp=e[g[x][i]];
             if(d[tmp.to]==d[x]+&&(f=dfs(tmp.to,min(a,tmp.c-tmp.f)))>){
                 tmp.f+=f;
                 e[g[x][i]^].f-=f;
                 flow+=f;
                 a-=f;
                 if(a==)break;
             }
         }
         if(!flow)d[x]=-;
         return flow;
     }

     int mf(int s,int t){
         this->s=s;
         this->t=t;
         int flow=;
         while(bfs()){
             memset(cur,,sizeof(cur));
             flow+=dfs(s,INF);
         }
         return flow;
     }
 };

 int num[];

 int main(){
     int n,m;
     while(scanf("%d%d",&n,&m)!=EOF){
         memset(num,,sizeof(num));
         dinic d;
         d.init(m+(<<m)+);
         int i,j;
         for(i=;i<=n;i++){
             int tmp=;
             for(j=;j<=m;j++){
                 int a=;
                 char c=getchar();
                 while(c>''||c<'')c=getchar();
                 a=c-'';
                 tmp=tmp*+a;
             }
             num[tmp]++;
         }
         for(i=;i<=(<<m)-;i++){
             if(num[i]){
                 d.add(,i,num[i]);
                 int tmp=i;
                 for(j=m;j>=;j--){
                     if(tmp&){
                         d.add(i,(<<m)+j,INF);
                     }
                     tmp>>=;
                 }
             }
         }
         for(j=;j<=m;j++){
             int a=;
             char c=getchar();
             while(c>''||c<'')c=getchar();
             while(c>=''&&c<=''){
                 a=a*+c-'';
                 c=getchar();
             }
             d.add((<<m)+j,m+(<<m)+,a);
         }
         if(n==d.mf(,m+(<<m)+))printf("YES\n");
         else printf("NO\n");
     }
     return ;
 }

最大流