BZOJ 1059 [ZJOI2007]矩阵游戏
题解:二分图匹配
行列建点
每一条边i->j表示第i行移动到第j行符合题意,即a[i][j]=1
可证明不需考虑列的交换
求最大匹配即可
#include<iostream> #include<cstdio> #include<cstring> #include<vector> #include<queue> using namespace std; const int maxn=300; const int oo=1000000000; int T; int n; struct Edge{ int from,to,cap,flow; }; vector<int>G[maxn*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; int vis[maxn*maxn]; int d[maxn*maxn]; queue<int>q; 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((e.cap>e.flow)&&(!vis[e.to])){ 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; 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 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(); while(!q.empty())q.pop(); } int main(){ scanf("%d",&T); while(T--){ Dinicinit(); scanf("%d",&n); s=n+n+1;t=s+1; for(int i=1;i<=n;++i){ Addedge(s,i,1); Addedge(i+n,t,1); } for(int i=1;i<=n;++i){ for(int j=1;j<=n;++j){ int x; scanf("%d",&x); if(x==1)Addedge(i,j+n,1); } } if(Maxflow()==n)printf("Yes\n"); else printf("No\n"); } return 0; }
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