poj2226——Muddy Fields

最小顶点覆盖=最大二分匹配。

只不过，这道题的构图比较麻烦，构图总的意思：扫描一遍输入的矩阵，把每行和每列中相邻的障碍物看成一个点，记录一下，然后加边。这样就转化成了求最大匹配的问题！见：http://www.cppblog.com/RyanWang/archive/2009/09/04/95283.aspx

#include<stdio.h>#include<string.h>#define max 555char g[max][max];int cx[max][max],cy[max][max],vis[max],my[max],gra[max][max];int r,c,nx,ny;bool find(int v){int j;for(j=1;j<=ny;j++){if(gra[v][j]&&vis[j]==0){vis[j]=1;if(my[j]==0||find(my[j])){my[j]=v;return true;}}}return false;}void solve(){int i;int ans=0;memset(my,0,sizeof(my));for(i=1;i<=nx;i++){memset(vis,0,sizeof(vis));if(find(i))ans++;}printf("%d/n",ans);}int main(){int i,j,k;scanf("%d%d",&r,&c);getchar();for(i=1;i<=r;i++){for(j=1;j<=c;j++)g[i][j]=getchar();getchar();}nx=0;ny=0;for(i=1;i<=r;i++){j=1;while(j<=c){if(g[i][j]=='*'){nx++;while(g[i][j]=='*'){cx[i][j]=nx;j++;}}else j++;}}for(j=1;j<=c;j++){i=1;while(i<=r){if(g[i][j]=='*'){ny++;while(g[i][j]=='*'){cy[i][j]=ny;i++;}}else i++;}}memset(gra,0,sizeof(gra));for(i=1;i<=r;i++)for(j=1;j<=c;j++){if(g[i][j]=='*')gra[cx[i][j]][cy[i][j]]=1;}solve();return 0;}