HackerRank - organizing-containers-of-balls

对于每种球，一定要有一个箱子所容纳的球的数量等于该类球的数量，才能保证同种球都在一个箱子里且该箱子中没有其他球

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <queue>
#include <vector>
#include <math.h>
#include <stack>
#include <map>
using namespace std;
const int MAX = 100+10;
const double eps = 1e-10;
const double PI = acos(-1.0);
int n, t, m[MAX][MAX];
bool vis[MAX], vis2[MAX];
bool check()
{
    for(int i = 0; i<n; ++i)
        if(!vis[i]||!vis2[i])return 0;
    return 1;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d", &n);
        memset(m,0,sizeof(m));
        for(int i = 0; i<n; ++i)
            for(int j = 0; j<n; ++j)
                {
                    scanf("%d", &m[i][j]);
                    m[n][i]+=m[i][j];
                    m[j][n]+=m[i][j];
                }
        memset(vis, 0, sizeof(vis));
        memset(vis2, 0, sizeof(vis2));
        for(int i = 0; i<n; ++i)
        {
            if(vis[i])continue;
            for(int j = 0; j<n; ++j)
            {
                if(vis2[j])continue;
                if(m[n][i]==m[j][n]&&!vis[i]&&!vis2[j])
                {
                    vis[i] = 1;
                    vis2[j] = 1;
                    break;
                }
            }
        }
        if(check())printf("Possible\n");
        else printf("Impossible\n");
    }
    return 0;
}