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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
#346538#4513. Slide Paradeorz_z35 ✓2522ms49672kbC++142.8kb2024-03-08 17:28:572024-03-08 17:28:57

Judging History

你现在查看的是最新测评结果

  • [2024-03-08 17:28:57]
  • 评测
  • 测评结果:35
  • 用时:2522ms
  • 内存:49672kb
  • [2024-03-08 17:28:57]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
const int N = 5e2 + 5, M = 4e4 + 5;
int n, m;
int U[M], V[M];
int fir[N], nxt[M], st[M], to[M], ect = 1;
inline void addedge(int u1, int v1) {
	nxt[++ect] = fir[u1];
	fir[u1] = ect;
	to[ect] = v1;
	st[ect] = u1;
}
bool vis[N];
int mth[N], mthe[N];
bool dfs(int x) {
	for (int i = fir[x], y; y = to[i], i; i = nxt[i])
		if (!vis[y]) {
			vis[y] = 1;
			if (!mth[y] || dfs(mth[y])) {
				mth[x] = y;
				mth[y] = x;
				mthe[x] = i;
				return 1;
			}
		}
	return 0;
}
int pv[N], pe[N];
void Bfs(int st) {
	queue<int> Q;
	for (int i = 1; i <= n + n; i++) vis[i] = pv[i] = pe[i] = 0;
	Q.push(st);
	vis[st] = true;
	while (!Q.empty()) {
		int x = Q.front();
		Q.pop();
		// printf("st,x:%d,%d\n",st,x);
		if (x > n) {
			int y = mth[x];
			if (vis[y]) continue;
			pv[y] = x;
			pe[y] = mthe[y];
			Q.push(y);
			vis[y] = true;
		} else
			for (int i = fir[x], y; y = to[i], i; i = nxt[i])
				if (i != mthe[x] && !vis[y]) {
					vis[y] = true;
					pv[y] = x;
					pe[y] = i;
					Q.push(y);
				}
	}
}
vector<int> Ans;
namespace Euler {
	const int N = 1e3 + 5, M = 4e6 + 5;
	int fir[N], nxt[M], to[M], ect = 1, sgn[M];
	inline void addvir(int u1, int v1) {
		// printf("vir:%d->%d\n",u1,v1);
		nxt[++ect] = fir[u1];
		fir[u1] = ect;
		to[ect] = v1;
		sgn[ect] = 0;
	}
	void Euler(int x) {
		for (int i = fir[x], y; y = to[i], i; i = fir[x]) {
			fir[x] = nxt[i];
			if (sgn[i]) continue;
			sgn[i] =  true;
			Euler(y);
		}
		Ans.push_back(x);
	}
	void Clr(int n) {
		ect = 1;
		for (int i = 1; i <= n; i++) fir[i] = 0;
		Ans.clear();
	}
}
bool inmth[M];
void work() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) cin >> U[i] >> V[i];
	for (int i = 1; i <= n + n; i++)
		pv[i] = pe[i] = mth[i] = vis[i] = mthe[i] = fir[i] = 0;
	Euler::Clr(n);
	ect = 1;
	for (int i = 1; i <= m; i++)
		addedge(U[i], V[i] + n);
	for (int i = 1; i <= n; i++) if(!mth[i]) {
		for (int j = 1; j <= n + n; j++) vis[j] = 0;
		if (!dfs(i)) return puts("IMPOSSIBLE"), void();
	}
	for (int i = 1; i <= ect; i++) inmth[i] = 0;
	for (int i = 1; i <= n; i++) inmth[mthe[i]] = true;
	for (int i = 1; i <= m; i++) {
		int u = U[i], v = V[i] + n;
		if (mth[u] != v) {
			mth[mth[u]] = 0;
			for (int j = 1; j <= n + n; j++) vis[j] = 0;
			vis[v] = 1;
			if (!dfs(mth[v])) return puts("IMPOSSIBLE"), void();
			else mth[u] = v, mth[v] = u;
		}
		for (int j = 1; j <= n; j++)
			Euler::addvir(j, mth[j] - n);
	}
	Euler::Euler(1);
	if (Ans.size() != Euler::ect) return puts("IMPOSSIBLE"), void();
	printf("%d\n", (int)Ans.size());
	reverse(Ans.begin(), Ans.end());
	// Ans.push_back(1);
	for (auto i : Ans) printf("%d ", i);
	printf("\n");
}
int main() {
	int T;
	cin >> T;
	for (int _ = 1; _ <= T; _++) {
		printf("Case #%d: ", _);
		work();
	}
	return 0;
}

详细

Test #1:

score: 11
Accepted
time: 2ms
memory: 10052kb

input:

100
5 8
1 2
1 3
1 4
1 5
2 1
3 1
4 1
5 1
5 10
1 3
1 4
2 3
2 5
3 1
3 4
3 5
4 2
5 1
5 3
5 10
1 4
2 3
2 5
3 1
3 5
4 2
4 3
4 5
5 1
5 2
3 6
1 2
1 3
2 1
2 3
3 1
3 2
5 10
1 2
1 5
2 3
2 4
3 1
4 3
4 5
5 2
5 3
5 4
4 10
1 2
1 3
1 4
2 1
2 3
2 4
3 1
3 4
4 2
4 3
5 10
1 2
1 3
2 1
2 4
3 1
3 5
4 2
4 5
5 3
5 4
5 10
1 ...

output:

Case #1: IMPOSSIBLE
Case #2: 51
1 4 2 5 3 1 4 2 3 5 1 4 2 3 5 1 4 2 3 5 1 3 4 2 5 1 4 2 5 3 1 3 4 2 5 1 4 2 3 5 1 4 2 3 5 1 3 4 2 5 1 
Case #3: 51
1 4 5 2 3 1 4 2 3 5 1 4 5 2 3 1 4 3 1 4 2 5 2 3 5 1 4 2 3 5 1 4 3 1 4 3 1 4 5 2 5 2 5 2 3 1 4 5 2 3 1 
Case #4: 19
1 3 2 1 2 3 1 2 3 1 3 2 1 3 2 1 2 3 1 ...

result:

ok  (100 test cases)

Test #2:

score: 24
Accepted
time: 2522ms
memory: 49672kb

input:

100
199 4980
1 95
1 96
1 105
1 124
1 130
1 131
1 135
1 137
1 139
1 140
1 141
1 147
1 155
1 172
1 174
1 179
1 183
1 188
1 193
1 196
1 197
2 3
2 5
2 13
2 14
2 17
2 20
2 24
2 26
2 30
2 41
2 44
2 45
2 52
2 56
2 67
2 70
2 71
2 74
2 78
2 84
2 85
2 90
2 92
2 93
2 97
2 107
2 111
2 113
2 122
2 124
2 128
2 13...

output:

Case #1: IMPOSSIBLE
Case #2: IMPOSSIBLE
Case #3: 1000001
1 193 80 66 24 105 108 168 90 64 13 28 199 176 70 129 114 65 97 92 15 151 34 62 43 38 53 82 175 25 136 47 147 161 112 123 57 8 37 22 16 50 166 127 156 20 42 149 11 73 140 145 146 125 137 99 14 101 17 52 157 138 200 169 191 102 154 148 159 122 ...

result:

ok  (100 test cases)