結果

問題 No.1775 Love Triangle 2
ユーザー 👑 ygussany
提出日時 2021-11-19 18:47:31
言語 C
(gcc 13.3.0)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 2,797 bytes
コンパイル時間 274 ms
コンパイル使用メモリ 32,640 KB
実行使用メモリ 8,736 KB
最終ジャッジ日時 2024-07-03 21:08:03
合計ジャッジ時間 9,775 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
other AC * 4 TLE * 1 -- * 85
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ソースコード

diff #

#include <stdio.h>

#define N_MAX 100
#define M_MAX 5000



typedef struct Edge {
	struct Edge *next;
	int v;
	unsigned int label;
} edge;

void DFS_naive(edge* adj[], int X, int Y, int Z, int u, int flag[], int* tmp, int* ans)
{
	if (flag[u] != 0) {
		if (u == X && flag[Y] != 0 && flag[Z] != 0) *ans = *tmp;
		return;
	} else flag[u] = 1;
	
	edge *p;
	(*tmp)++;
	if (*tmp < *ans) for (p = adj[u]; p != NULL; p = p->next) DFS_naive(adj, X, Y, Z, p->v, flag, tmp, ans);
	(*tmp)--;
	flag[u] = 0;
}

// Naive solution
int naive(int N, int M, int X, int Y, int Z, int A[], int B[])
{
	static int i, u, w, adj_mat[N_MAX + 1][N_MAX + 1];
	static edge *adj[N_MAX + 1], e[M_MAX * 2], *p;
	for (u = 1; u <= N; u++) for (w = u + 1; w <= N; w++) adj_mat[u][w] = 0;
	for (i = 1; i <= M; i++) {
		u = A[i];
		w = B[i];
		adj_mat[u][w] = 1;
	}
	for (u = 1; u <= N; u++) adj[u] = NULL;
	for (u = 1, i = 0; u <= N; u++) {
		for (w = u + 1; w <= N; w++) {
			if (adj_mat[u][w] != 0) continue;
			e[i].v = w;
			e[i].next = adj[u];
			adj[u] = &(e[i++]);
			e[i].v = u;
			e[i].next = adj[w];
			adj[w] = &(e[i++]);
		}
	}
	
	static int flag[N_MAX + 1] = {}, q[N_MAX + 1], head, tail;
	
	// Y--Z path?
	flag[X] = 1;
	flag[Y] = 1;
	q[0] = X;
	q[1] = Y;
	for (head = 1, tail = 2; head < tail; head++) {
		u = q[head];
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			if (flag[w] == 0) {
				flag[w] = 1;
				q[tail++] = w;
			}
		}
	}
	if (flag[Z] == 0) {
		for (head = 0; head < tail; head++) flag[q[head]] = 0;
		return -1;
	} else for (head = 0; head < tail; head++) flag[q[head]] = 0;
	
	// Z--X path?
	flag[Y] = 1;
	flag[Z] = 1;
	q[0] = Y;
	q[1] = Z;
	for (head = 1, tail = 2; head < tail; head++) {
		u = q[head];
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			if (flag[w] == 0) {
				flag[w] = 1;
				q[tail++] = w;
			}
		}
	}
	if (flag[X] == 0) {
		for (head = 0; head < tail; head++) flag[q[head]] = 0;
		return -1;
	} else for (head = 0; head < tail; head++) flag[q[head]] = 0;
	
	// X--Y path?
	flag[Z] = 1;
	flag[X] = 1;
	q[0] = Z;
	q[1] = X;
	for (head = 1, tail = 2; head < tail; head++) {
		u = q[head];
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			if (flag[w] == 0) {
				flag[w] = 1;
				q[tail++] = w;
			}
		}
	}
	if (flag[Y] == 0) {
		for (head = 0; head < tail; head++) flag[q[head]] = 0;
		return -1;
	} else for (head = 0; head < tail; head++) flag[q[head]] = 0;
	
	// DFS
	int ans = N + 1, tmp = 0;
	DFS_naive(adj, X, Y, Z, X, flag, &tmp, &ans);
	if (ans > N) return -1;
	else return ans;
}



int main()
{
	int i, N, M, X, Y, Z, A[M_MAX + 1], B[M_MAX + 1];
	scanf("%d %d", &N, &M);
	scanf("%d %d %d", &X, &Y, &Z);
	for (i = 1; i <= M; i++) scanf("%d %d", &(A[i]), &(B[i]));
	printf("%d\n", naive(N, M, X, Y, Z, A, B));
	fflush(stdout);
	return 0;
}
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