USACO 3.3 Home on the Range (range)

/*Main idea这道题可以动态规划。二维的动态规划。状态定义：G[i][j]为以(i,j)为左上角顶点的正方形的最大边长。边界条件：G[i][j]为初始读入的矩阵。状态转移方程： G[i][j]=min{ G[i+1][j] , G[i][j+1] , G[i+1][j+1] } + 1;解析： G[i+1][j] , G[i][j+1] , G[i+1][j+1]分别为(i,j)向下、向右、向右下一格的状况。在(n-1,n-1)当且仅当三者都为1的时候，正方形才能扩充。从最右下向上，依次扩充即可。refer to byvoid;Note Dynamic programming's way to count the number of squares for each size;This is a way of collecting solution during DP;*//*Executing...   Test 1: TEST OK [0.000 secs, 3748 KB]   Test 2: TEST OK [0.000 secs, 3748 KB]   Test 3: TEST OK [0.000 secs, 3748 KB]   Test 4: TEST OK [0.000 secs, 3748 KB]   Test 5: TEST OK [0.000 secs, 3748 KB]   Test 6: TEST OK [0.011 secs, 3748 KB]   Test 7: TEST OK [0.054 secs, 3748 KB]All tests OK.*//*ID: haolink1PROG: rangeLANG: C++*///#include <iostream>#include <fstream>#include <string>using namespace std;const int MAX = 250;int dp[MAX][MAX];int ans[MAX+1];int N = 0;ifstream fin("range.in");ofstream cout("range.out");void Init(){    fin >> N;    string str;    for(int i = 0; i < N; i++){        fin >> str;        for(int j = 0; j < N; j++){            dp[i][j] = str[j] - 48;        }    }}int Min(int a,int b,int c){    int min = a <= b ? a : b;    min = min <= c ? min : c;    return min;}void Dynamic(){    for(int i = N - 2; i >= 0; i--){        for(int j = N - 2; j >= 0; j--){            if(dp[i][j])//Note dp[i][j] must first be 1;                dp[i][j] = Min(dp[i+1][j],dp[i+1][j+1],dp[i][j+1]) + 1;            if(dp[i][j] > 1){                for(int k = 2; k <= dp[i][j]; k++){//count the size of square whose upleft point is(i,j);                    ans[k]++;                }            }        }    }}void Print(){    for(int i = 2; i <= N; i++){        if(ans[i]){            cout << i <<" "<< ans[i] << endl;         }    }}int main(){    Init();     Dynamic();    Print();    return 0;}