usaco Home on the Range

dp关系式搞清楚就行了，我看题解的没想出来。a[i][j]=min(a[i+1][j],a[i][j+1],a[i+1][j+1])+1;他的右方下方右下方保证了他可以延伸的长度。

/*ID: jinbo wuTASK: rangeLANG:C++*/#include<bits/stdc++.h>using namespace std;int a[300][300];int cnt[300];int main(){freopen("range.in","r",stdin);freopen("range.out","w",stdout);int n;scanf("%d",&n);for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){scanf("%1d",&a[i][j]);}for(int i=n-1;i>=1;i--)for(int j=n-1;j>=1;j--){if(a[i][j])a[i][j]=min(a[i+1][j],min(a[i][j+1],a[i+1][j+1]))+1;}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){while(a[i][j]>=2)//大正方形包括了小正方形{cnt[a[i][j]]++;a[i][j]--;}}for(int i=2;i<=n;i++){if(cnt[i]){printf("%d %d\n",i,cnt[i]);}}}