题解:二分图建模
左边是人,右边是床
s向需要在学校的人连边
有床的人向t连边
认识的人互相连边
跑最大流与需要在学校的人数量是否相等比较、
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int maxn=6000;
const int oo=1000000000;
int TT;
int n;
int isstu[100];
int isgoh[100];
struct Edge{
int from,to,cap,flow;
};
vector<int>G[maxn];
vector<Edge>edges;
void Addedge(int x,int y,int z){
Edge e;
e.from=x;e.to=y;e.cap=z;e.flow=0;
edges.push_back(e);
e.from=y;e.to=x;e.cap=0;e.flow=0;
edges.push_back(e);
int c=edges.size();
G[x].push_back(c-2);
G[y].push_back(c-1);
}
int s,t;
queue<int>q;
int vis[maxn];
int d[maxn];
int Bfs(){
memset(vis,0,sizeof(vis));
vis[s]=1;d[s]=0;q.push(s);
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])&&(e.cap>e.flow)){
vis[e.to]=1;
d[e.to]=d[x]+1;
q.push(e.to);
}
}
}
return vis[t];
}
int Dfs(int x,int a){
if((x==t)||(a==0))return a;
int nowflow=0,f=0;
for(int i=0;i<G[x].size();++i){
Edge e=edges[G[x][i]];
if((d[x]+1==d[e.to])&&((f=Dfs(e.to,min(a,e.cap-e.flow)))>0)){
nowflow+=f;
a-=f;
edges[G[x][i]].flow+=f;
edges[G[x][i]^1].flow-=f;
if(a==0)break;
}
}
return nowflow;
}
int ma[100][100];
int Maxflow(){
int flow=0;
while(Bfs())flow+=Dfs(s,oo);
return flow;
}
void Dinicinit(){
for(int i=0;i<=n+n+2;++i)G[i].clear();
edges.clear();
}
int main(){
scanf("%d",&TT);
while(TT--){
scanf("%d",&n);
Dinicinit();
int cnt=0;
for(int i=1;i<=n;++i)scanf("%d",&isstu[i]);
for(int i=1;i<=n;++i)scanf("%d",&isgoh[i]);
//n+n
s=n+n+1;t=s+1;
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j){
scanf("%d",&ma[i][j]);
}
}
for(int i=1;i<=n;++i){
if((isstu[i]&&!isgoh[i])||(!isstu[i])){
++cnt;
Addedge(s,i,1);
}
}
for(int i=1;i<=n;++i){
if(isstu[i])Addedge(i+n,t,1);
}
for(int i=1;i<=n;++i){
for(int j=1;j<=n;++j){
if(ma[i][j]){
Addedge(i,j+n,1);
}
}
}
for(int i=1;i<=n;++i)Addedge(i,i+n,1);
if(Maxflow()!=cnt)printf("T_T
");
else printf("^_^
");
}
return 0;
}