lp2356

https://www.luogu.org/problemnew/show/P2356

一维数组解决二维题目，妙啊

#include<cstdio>
#include<iostream>
using namespace std;
int x[10005]={0},y[10005]={0},n,m,count = 0;
//x,y 是每个"容身之地"的坐标，count为其个数 
int flag = 0;// 记录有没有"容身之地"
int maxn = -1;
int main(){
    int hang[1005]={0},lie[1005]={0}; 
    //保存每行每列的和 
    cin >> n;
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= n;j++){
            scanf("%d",&m);
            hang[i] += m; 
            lie[j] += m; //把值加上去 
            if(m == 0){  //有"容身之地" 
                count++;//"容身之地"个数加一
                x[count] = i;
                y[count] = j; //保存x y 
                flag = 1;
            }
        }
    }
    if(flag == 0) printf("Bad Game!\n");//没有"容身之地"
    if(flag == 1){
        for(int i = 1;i <= count;i++){
            if(hang[x[i]] + lie[y[i]] > maxn)
                maxn = hang[x[i]] + lie[y[i]]; //计算每个没有"容身之地"的行列之和 
        }
        cout << maxn << endl;
    }
    return 0;
}