QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#612076#6705. Mediankevinyang#AC ✓3ms3712kbC++143.2kb2024-10-05 04:26:282024-10-05 04:26:29

Judging History

你现在查看的是最新测评结果

  • [2024-10-05 04:26:29]
  • 评测
  • 测评结果:AC
  • 用时:3ms
  • 内存:3712kb
  • [2024-10-05 04:26:28]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define int long long

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

bool check(vector<vector<int>>adj, int n){
    vector<bool>vis(n+1);
    queue<int>q;
    vector<int>deg(n+1);
    for(int i = 1; i<=n; i++){
        for(int j: adj[i]){
            deg[j]++;
        }
    }
    for(int i = 1; i<=n; i++){
        if(deg[i] == 0){
            q.push(i);
        }
    }
    while(q.size()){
        int cur = q.front(); q.pop();
        vis[cur] = 1;
        for(int nxt: adj[cur]){
            deg[nxt]--;
            if(deg[nxt] == 0){
                q.push(nxt);
            }
        }
    }
    return count(vis.begin(),vis.end(),1) == n;
}

signed main() {
    cin.tie(0)->sync_with_stdio(0);
    int t;
    cin >> t;
    while(t--){
        int n,m;
        cin >> n >> m;
        vector<vector<int>>adj(n+1);
        vector<vector<int>>rev(n+1);
        for(int i = 0; i<m; i++){
            int x,y;
            cin >> x >> y;
            adj[x].push_back(y);
            rev[y].push_back(x);
        }
        if(!check(adj,n)){
            for(int i = 0; i<n; i++){
                cout << "0";
            }
            cout << '\n';
            continue;
        }
        vector<bitset<105>>dp(n+1);
        vector<bitset<105>>dp2(n+1);

        {
            queue<int>q;
            vector<int>deg(n+1);
            for(int i = 1; i<=n; i++){
                for(int j: adj[i]){
                    deg[j]++;
                }
            }
            for(int i = 1; i<=n; i++){
                if(deg[i] == 0){
                    q.push(i);
                }
                dp[i][i] = 1;
            }
            while(q.size()){
                int cur = q.front(); q.pop();
                for(int nxt: adj[cur]){
                    deg[nxt]--;
                    dp[nxt] |= dp[cur];
                    if(deg[nxt] == 0){
                        q.push(nxt);
                    }
                }
            }
        }
        {
            queue<int>q;
            vector<int>deg(n+1);
            for(int i = 1; i<=n; i++){
                for(int j: rev[i]){
                    deg[j]++;
                }
            }
            for(int i = 1; i<=n; i++){
                if(deg[i] == 0){
                    q.push(i);
                }
                dp2[i][i] = 1;
            }
            while(q.size()){
                int cur = q.front(); q.pop();
                for(int nxt: rev[cur]){
                    deg[nxt]--;
                    dp2[nxt] |= dp2[cur];
                    if(deg[nxt] == 0){
                        q.push(nxt);
                    }
                }
            }
        }
        for(int i = 1; i<=n; i++){
            int c1 = dp[i].count();
            int c2 = dp2[i].count();
            if(c1 <= (n+1)/2 && c2 <= (n+1)/2){
                cout << 1;
            }
            else{
                cout << 0;
            }
        }
        cout << '\n';
    }
    return 0;
}

这程序好像有点Bug,我给组数据试试?

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3608kb

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: 3ms
memory: 3712kb

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