水题 Codeforces Round #303 (Div. 2) A. Toy Cars

题目传送门

/*
    题意：5种情况对应对应第i或j辆车翻了没
    水题：其实就看对角线的上半边就可以了，vis判断，可惜WA了一次
    3: if both cars turned over during the collision.
    是指i，j两辆车，而不是全部
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
using namespace std;

const int MAXN = 1e2 + 10;
const int INF = 0x3f3f3f3f;
int a[MAXN][MAXN];
bool vis[MAXN];
int ans[MAXN];

int main(void)        //Codeforces Round #303 (Div. 2) A. Toy Cars
{
    //freopen ("A.in", "r", stdin);

    int n;
    while (scanf ("%d", &n) == 1)
    {
        memset (vis, false, sizeof (vis));
        for (int i=1; i<=n; ++i)
        {
            for (int j=1; j<=n; ++j)    scanf ("%d", &a[i][j]);
        }

        for (int i=1; i<=n; ++i)
        {
            for (int j=i+1; j<=n; ++j)
            {
                if (a[i][j] == 3)
                {
                    vis[i] = vis[j] = true;
                }
                else if (a[i][j] == 2)    vis[j] = true;
                else if (a[i][j] == 1)    vis[i] = true;
            }
        }

        int cnt = 0;
        for (int i=1; i<=n; ++i)
        {
            if (!vis[i])    ans[++cnt] = i;
        }
        printf ("%d
", cnt);
        for (int i=1; i<=cnt; ++i)    printf ("%d%c", ans[i], (i==cnt) ? '
' : ' ');
    }

    return 0;
}