题解:
对于每一联通的x,y
检点
然后交叉的连边
然后二分图
代码:
#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const int N=1250; int cas,n,m,num,t,x[N][N],y[N][N],f[N],fi[N],ne[N],zz[N],match[N],r,c; char s[N][N]; void jb(int x,int y) { zz[++num]=y; ne[num]=fi[x]; fi[x]=num; } void build() { r=1,c=1; for (int i=0;i<n;i++) for (int j=0;j<m;j++) { if (s[i][j]!='#'&&!x[i][j]) { for (int k=j;s[i][k]!='#'&&k<m;k++){x[i][k]=r;} r++; } if (s[i][j]!='#'&&!y[i][j]) { for (int k=i;s[k][j]!='#'&&k<n;k++){y[k][j]=c;} c++; } } for (int i=0;i<n;i++) for (int j=0;j<m;j++) if (s[i][j]=='o'&&x[i][j]&&y[i][j])jb(x[i][j],y[i][j]); } int dfs(int x) { for (int i=fi[x];i;i=ne[i]) if (!f[zz[i]]) { f[zz[i]]=1; if (!match[zz[i]]||dfs(match[zz[i]])) { match[zz[i]]=x; return 1; } } return 0; } int main() { scanf("%d",&t); while (t--) { memset(x,0,sizeof x); memset(y,0,sizeof y); memset(fi,0,sizeof fi); memset(match,0,sizeof match); num=0; scanf("%d%d",&n,&m); for (int i=0;i<n;i++)scanf("%s",s[i]); build(); printf("Case :%d ",++cas); int ans=0; for (int i=1;i<=r;i++) { memset(f,0,sizeof f); ans+=dfs(i); } printf("%d ",ans); } }