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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
#77683#5511. Minor EvilXKErrorWA 4ms27116kbC++2.0kb2023-02-15 12:46:402023-02-15 12:46:43

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-02-15 12:46:43]
  • 评测
  • 测评结果:WA
  • 用时:4ms
  • 内存:27116kb
  • [2023-02-15 12:46:40]
  • 提交

answer

#include <bits/stdc++.h>

#define maxn 1000005

using namespace std;

int n, k;
int a[maxn];
int b[maxn];

int m;
int s[maxn];

vector<pair<int, int> > g[maxn];

void add(int u, int v, int w) {
//	cout<<"ADd:"<<u<<" "<<v<<" "<<w<<endl;
	g[u].push_back({v, w});
//	g[v].push_back({u, w});
}

int f[maxn];
int res[maxn];

int alv[maxn];

queue<int> q;

int main() {
//	freopen("dt.in", "r", stdin);
	int T;
	scanf("%d", &T);
	while (T--) {
		scanf("%d%d", &n, &k);
		for (int i = 1; i <= n; i++) f[i] = k + 1;
		for (int i = 1; i <= k; i++) scanf("%d%d", &a[i], &b[i]);
		scanf("%d", &m);
		for (int i = 1; i <= m; i++) {
			int x;
			scanf("%d", &x);
			s[x] = 1;
		}
		for (int i = 1; i <= k; i++) {
			if (s[b[i]]) add(a[i], b[i], i);
		}
		for (int i = 1; i <= n; i++) if (!s[i]) q.push(i);
		while (!q.empty()) {
			int now = q.front(); q.pop();
			for (auto p : g[now]) {
				int v = p.first, w = p.second;
//				cout<<now<<" "<<f[now]<<" "<<w<<" "<<f[v]<<endl;
				if (f[now] > w && (f[v] < w || f[v] == k + 1)) {
					f[v] = w;
					q.push(v);
				}
			}
		}
		bool flg = 1;
		for (int i = 1; i <= n; i++) {
			if (s[i]) {
//				cout<<f[i]<<"@"<<endl;
				if (f[i] <= k) res[f[i]] = 1;
				else flg = 0;
			}
		}
		if (!flg) {
			puts("NIE");
			for (int i = 1; i <= k; i++) res[i] = f[i] = s[i] = 0, g[i].clear();
			continue;
		}
		puts("TAK");
		for (int i = 1; i <= k; i++) putchar(res[i] ? 'T' : 'N');
		puts("");
		for (int i = 1; i <= n; i++) alv[i] = 1;
		for (int i = 1; i <= k; i++) {
			if (res[i]) {
				if (alv[a[i]] && alv[b[i]]) alv[b[i]] = 0;
			}
		}
		bool fl = 1;
		for (int i = 1; i <= n; i++) fl &= (!s[i] || !alv[i]);
		if (!fl) {
			printf("%d %d\n", n, k);
			for (int i = 1; i <= k; i++) printf("%d %d\n", a[i], b[i]);
			printf("%d\n", m);
			for (int i = 1; i <= n; i++) if (s[i]) printf("%d ", i);
			puts("");
			return 0;
		}
		for (int i = 1; i <= k; i++) res[i] = f[i] = s[i] = 0, g[i].clear();
	}
	return 0;
}

详细

Test #1:

score: 100
Accepted
time: 3ms
memory: 26924kb

input:

2
5 6
1 2
2 1
2 5
2 3
2 4
4 2
3
1 2 3
3 2
1 2
2 3
2
2 3

output:

TAK
NTNTNT
NIE

result:

ok correct (2 test cases)

Test #2:

score: -100
Wrong Answer
time: 4ms
memory: 27116kb

input:

1000
5 6
1 2
2 1
2 5
2 3
2 4
4 2
3
1 2 3
3 2
1 2
2 3
2
2 3
2 1
1 2
1
1
2 1
1 2
1
2
3 3
2 1
3 2
3 2
1
3
3 3
1 3
1 3
1 2
2
1 3
3 3
1 2
1 3
1 3
1
2
3 3
2 1
2 3
1 3
1
2
3 3
3 2
3 1
1 2
3
1 2 3
3 3
1 2
2 3
1 2
1
3
3 3
2 1
1 2
1 2
1
2
3 3
2 1
1 3
1 3
1
1
3 3
3 2
3 2
2 3
1
3
3 3
3 2
1 2
2 1
1
1
3 3
2 1
3 2...

output:

TAK
NTNTNT
NIE
NIE
TAK
T
NIE
NIE
TAK
TNN
NIE
NIE
TAK
NTN
TAK
NNT
TAK
TNN
TAK
NNT
TAK
NNT
TAK
NNT
NIE
NIE
NIE
NIE
NIE
NIE
NIE
NIE
NIE
TAK
NTNN
TAK
TNTN
NIE
NIE
NIE
NIE
NIE
NIE
NIE
TAK
TNTN
TAK
NNTN
TAK
NNNT
TAK
NNTN
NIE
TAK
NNTN
NIE
NIE
TAK
NNNT
NIE
TAK
NNTN
NIE
NIE
NIE
NIE
NIE
NIE
NIE
NIE
NIE
NIE
TA...

result:

wrong answer Contestant didn't find the solution. (test case 53)