算法竞赛（刘汝佳） 习题4-2 正方形 UVa201

Question

#include <iostream>
#include <cstdio>
using namespace std;
int n,m,w = 0, turn = 1;

int main(){
    while ( scanf("%d %d",&n,&m)==2 ){
        int arr_H[10][10]={{0,0}}, arr_V[10][10]={{0,0}};
        //存入数据
        for ( int q=0; q<m; q++ ){
            char ch;
            int i,j;
            cin >> ch >> i >> j;
            if ( ch=='H' ) arr_H[i-1][j-1] = 1;
            else arr_V[j-1][i-1] = 1;
        }
        
        //以边长遍历
        int count[10]={0}, ist = 0;
        for ( int j=1; j<=n; j++ ){
            ist ++;
            //以位置遍历
            for ( int x=0; x<n-j; x++ ){
                for ( int y=0; y<n-j; y++ ){
                    //判断是否是正方形
                    int judge = 1;
                    for ( int k=0; k<=j; k++ )
                        if ( !arr_H[x+k][y] || !arr_V[x][y+k] ) judge = 0;
                    if ( judge ) count[ist]++;
                }
            }
        }
        if ( w ) printf("\n**********************************\n\n");
        printf("Problem #%d\n\n",turn++);
        int judge = 1;
        for ( int i=1; i<=n; i++ ){
            if ( count[i]!=0 ){
                printf("%d square (s) of size %d\n",count[i],i);
                judge = 0;
            }
        }
        if ( judge ) printf("No completed squares can be found.\n");
        w = 1;
    }
    return 0;
}

整体思路是用两个数组，arr_H[] arr_V[]来保存横线和竖线，遍历每一个正方形的左上角顶点。count数组表示每个尺寸正方形的数量。我拿测试样例和自己想的样例，用别人ac代码和我的代码运行，结果都一样，但我的提交显示wa

Answer 1

看了一下,wa应该是你两个for循环思路不对,再好好想想吧233 或者看看教参