最大子矩陣和變形（poj 1964 hdu 1505)

解釋詳見：訓練指南P50

#include <iostream>#include <algorithm>using namespace std;int up[1005][1005],lef[1005][1005],righ[1005][1005];char a[1005][1005]; int main(){int t,m,n,ans,lo,ro,i,j;char s;cin>>t;while(t--){cin>>m>>n;for(i=0;i<m;i++)for(j=0;j<n;j++)cin>>a[i][j];ans=0;for(i=0;i<m;i++){   lo=-1;  ro=n;for(j=0;j<n;j++) {if(a[i][j]=='R') {lef[i][j]=up[i][j]=0; lo=j;}else {up[i][j]=(i==0)?1:up[i-1][j]+1;lef[i][j]=(i==0)?lo+1:max(lef[i-1][j],lo+1);}}for(j=n-1;j>=0;j--){if(a[i][j]=='R') {righ[i][j]=n; ro=j;}else  righ[i][j]=(i==0)?ro-1:min(righ[i-1][j],ro-1);ans=max(ans,up[i][j]*(righ[i][j]-lef[i][j]+1));}}cout<<ans*3<<endl;}return 0;}