poj3020——Antenna Placement

这道题，不分左右的x,y点，直接匹配。将最终的匹配数ans除以2即可。而答案的话，应该是result=cnt(总的点数)-2*(ans/2)+ans/2。

#include<iostream>#include<string>#include<cstdio>using namespace std;#define H 45#define maxn 450string date[H];int dir[][2]={{-1,0},{1,0},{0,-1},{0,1}};int map[H][H];int h,w;class node{public:int to;int next;};node g[maxn*100];int head[maxn],cnt,len;bool vis[maxn];int my[maxn];bool dfs(int v){int i;for(i=head[v];i;i=g[i].next ){int u=g[i].to ;if(vis[u]){vis[u]=false ;if(my[u]==-1||dfs(my[u])){my[u]=v;return true;}}}return false;}void solve(){int i,j,ans=0;for(i=1;i<=cnt;i++)my[i]=-1; for(i=1;i<=cnt;i++){memset(vis,true,sizeof(vis));if(dfs(i)) ans++;}ans/=2;int result=cnt-ans;cout<<result<<endl;}int main(){int t,i,j;cin>>t;while(t--){cin>>h>>w;memset(head,0,sizeof(head));for(i=0;i<h;i++)cin>>date[i];cnt=0;for(i=0;i<h;i++){for(j=0;j<w;j++)if(date[i].at (j)=='*')map[i][j]=++cnt;}len=0;for(i=0;i<h;i++)for(j=0;j<w;j++)if(date[i].at (j)=='*'){for(int k=0;k<4;k++){int tempx=i+dir[k][0];int tempy=j+dir[k][1];if(tempx>=0&&tempx<h&&tempy>=0&&tempy<w&&date[tempx].at (tempy)=='*'){g[++len].to =map[tempx][tempy];g[len].next =head[map[i][j]];head[map[i][j]]=len;}}}solve();}return 0;}