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  • 洛谷P3355 骑士共存问题 二分图_网络流

    Code:

    #include<cstdio>
    #include<cstring>
    #include<queue>
    #include<vector>
    #include<algorithm>
    using namespace std;
    const int maxn=800004;
    const int INF=10000000;
    int A[205][205],B[205][205];
    int dy[]={-1,-2,-2,-1,1,2,2,1};
    int dx[]={-2,-1,1,2,-2,-1,1,2};
    # define  pb push_back
    int s,t;
    struct Edge{
    	int from,to,cap;
    	Edge(int u,int v,int c):from(u),to(v),cap(c) {}
    };
    struct Dicnic{
       vector<Edge>edges;
       vector<int>G[maxn];
       int d[maxn],vis[maxn],cur[maxn];
       queue<int>Q;
       void addedge(int u,int v,int c){
    	edges.pb(Edge(u,v,c));               //正向弧
    	edges.pb(Edge(v,u,0));               //反向弧
    	int m=edges.size();
    	G[u].pb(m-2);
    	G[v].pb(m-1);
       }
       int BFS()
       {
    	memset(vis,0,sizeof(vis));
    	d[s]=0,vis[s]=1;Q.push(s);
    	while(!Q.empty()){
    		int u=Q.front();Q.pop();
    		int sz=G[u].size();
    		for(int i=0;i<sz;++i){
    			Edge e=edges[G[u][i]];
    			if(!vis[e.to]&&e.cap>0){
    				d[e.to]=d[u]+1,vis[e.to]=1;
    				Q.push(e.to);
    			}
    		}
    	}
    	return vis[t];
       }
       int dfs(int x,int a){
    	   if(x==t)return a;
    	   int sz=G[x].size();
    	   int f,flow=0;
    	   for(int i=cur[x];i<sz;++i){
    		Edge e=edges[G[x][i]];
    		cur[x]=i;
    		if(d[e.to]==d[x]+1&&e.cap>0){
    			f=dfs(e.to,min(a,e.cap));
    			if(f)
    			{
    				int u=G[x][i];
    				a-=f;
    				edges[u].cap-=f;
    				edges[u^1].cap+=f;
    				flow+=f;
    				if(a==0)break;
    			}
    		}
    	   }
    	   return flow;
       }
       int maxflow(){
    	int ans=0;
    	while(BFS()){
    	  memset(cur,0,sizeof(cur));
    	  ans+=dfs(s,INF);
    	}
    	return ans;
       }
    }op;
    int main()
    {
    	int n,m;
    	scanf("%d%d",&n,&m);
    	int cnt=0;
    	s=0;
    	for(int i=1;i<=n;++i)
    		for(int j=1;j<=n;++j){
                                         A[i][j]=++cnt;
    		}
    	t=cnt*2+1;
    	for(int i=1;i<=m;++i)
    	{
    		int a,b;
    		scanf("%d%d",&a,&b);
    		B[a][b]=1;
    	}
    	for(int i=1;i<=n;++i)
    	{
    		for(int j=1;j<=n;++j)
    		{
    			if(B[i][j])continue;
                                         for(int ii=0;ii<8;++ii)
                                         {
                                                  int yy=dy[ii]+i,xx=dx[ii]+j;
                                                          if(yy>=1&&yy<=n&&xx>=1&&xx<=n)
                                                          	 if(!B[yy][xx])op.addedge(A[i][j],A[yy][xx]+cnt,INF);
                                         }
    		}
    	}
    	for(int i=1;i<=n;++i)
    	        for(int j=1;j<=n;++j)
    	        {
    	    	if(!B[i][j])op.addedge(s,A[i][j],1);
    	    }
    	for(int i=cnt+1;i<=cnt+cnt;++i)
    	{
    		op.addedge(i,t,1);
    	}
    	int ans=cnt-op.maxflow()/2;
    	printf("%d
    ",ans-m);
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/guangheli/p/10367587.html
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