QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#410792#6705. Medianwzl19371AC ✓2ms3540kbC++141.1kb2024-05-14 15:01:162024-05-14 15:01:17

Judging History

你现在查看的是最新测评结果

  • [2024-05-14 15:01:17]
  • 评测
  • 测评结果:AC
  • 用时:2ms
  • 内存:3540kb
  • [2024-05-14 15:01:16]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
struct Edge{
    int to, nxt;
};
void solve(){
    int n,m; cin >> n >> m;
    vector<bitset<105>> f(n+2);
    for(int i=1;i<=m;i++) {
        int x,y; cin >> x >> y;
        f[x][y] = 1;
    }
    for(int k=1;k<=n;k++) {
        for(int i=1;i<=n;i++){
            if(f[i][k]) {
                f[i] |= f[k];
            }
        }
    }
    int pre[n+5]={}, suf[n+5]={};
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=n;j++) {
            if(f[i][j]) {    
                pre[i]++;
                suf[j]++;
            }
        }
    }
    for(int i=1;i<=n;i++) {
        if(f[i][i]) {
            cout << string(n, '0') << endl;
            return;
        }
    }
    for(int i=1;i<=n;i++) {
        if(pre[i] <= n/2 && suf[i] <= n/2) {
            cout << '1';
        }else {
            cout << '0';
        }
    }
    cout << endl;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t; cin >> t;
    while(t--)
        solve();
    return 0;
}

这程序好像有点Bug,我给组数据试试?

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3440kb

input:

2
5 4
1 2
3 2
2 4
2 5
3 2
1 1
2 3

output:

01000
000

result:

ok 2 lines

Test #2:

score: 0
Accepted
time: 2ms
memory: 3540kb

input:

66
13 2
9 13
7 11
11 19
9 1
8 1
5 1
2 8
4 2
2 1
5 2
6 3
3 11
3 2
4 6
6 10
9 8
3 5
1 7
5 8
3 9
4 9
6 7
3 1
2 3
11 6
9 4
1 6
5 2
1 5
4 6
8 4
15 15
10 6
15 8
7 6
11 1
5 2
3 4
11 13
4 6
10 12
10 13
1 6
15 2
5 12
13 14
5 3
15 86
14 12
8 1
14 9
8 15
5 10
1 9
11 2
6 2
7 10
10 13
14 5
4 13
5 8
4 10
13 9
6 9...

output:

1111111111111
01001000111
111
11111111111
111111111111111
001001000000000
00100
01100
1111111
1000000000000
111101101
111111111
000011111011101
010111111
001100000
0100001001101
1111111111111
001000010000000
10010111011
001000000000100
11111111111
00100000011
11111
01000000110
11101110111
00000
1111...

result:

ok 66 lines