結果

問題 No.1745 Selfish Spies 2 (à la Princess' Perfectionism)
ユーザー 👑 ygussanyygussany
提出日時 2021-11-07 20:00:26
言語 C
(gcc 13.3.0)
結果
TLE  
実行時間 -
コード長 4,022 bytes
コンパイル時間 211 ms
コンパイル使用メモリ 33,024 KB
実行使用メモリ 30,276 KB
最終ジャッジ日時 2024-11-30 04:59:29
合計ジャッジ時間 189,542 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
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ファイルパターン 結果
other AC * 31 TLE * 28
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ソースコード

diff # 

#include <stdio.h>
#include <stdlib.h>

#define N_MAX 100000
#define M_MAX 100000
#define L_MAX 200000

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



int DFS_bipartite_matching(edge* aux[], int par[], int u)
{
	int w;
	for (; aux[u] != NULL; aux[u] = aux[u]->next) {
		w = aux[u]->v;
		if (par[w] == 0) { // w is a sink
			par[w] = u;
			return w;
		} else if (par[w] > 0) continue; // w is already checked
		par[w] = u;
		w = DFS_bipartite_matching(aux, par, w);
		if (w > 0) return w;
	}
	return 0;
}

int bipartite_matching_augmentation(int N, char color[], edge* adj[], int mate[])
{
	static int i, u, w, depth[N_MAX + M_MAX + 1], par[N_MAX + M_MAX + 1], q[N_MAX + M_MAX + 1], head, tail;
	static edge *aux[N_MAX + M_MAX + 1], f[L_MAX * 2], *p;
	for (u = 1, tail = 0, par[0] = 0; u <= N; u++) {
		if (mate[u] == 0) { // u is a source of sink
			if (color[u] == 0) { // u is a source
				depth[u] = 0;
				q[tail++] = u;
			} else depth[u] = N;
			par[u] = 0;
		} else {
			depth[u] = N;
			par[u] = -1;
		}
	}
	
	// BFS for constructing the layered network
	for (head = 0, i = 0; head < tail; head++) {
		u = q[head];
		aux[u] = NULL;
		if (color[u] == 0) {
			for (p = adj[u]; p != NULL; p = p->next) {
				w = p->v;
				if (mate[u] == w) continue; // No arc in this direction
				if (depth[w] == N) { // w has been found for the first time
					depth[w] = depth[u] + 1;
					q[tail++] = w;
				}
				if (depth[w] == depth[u] + 1) { // Add the arc uw
					f[i].v = w;
					f[i].next = aux[u];
					aux[u] = &(f[i++]);
				}
			}
		} else if (mate[u] != 0) {
			w = mate[u];
			if (depth[w] == N) { // w has been found for the first time
				depth[w] = depth[u] + 1;
				q[tail++] = w;
			}
			if (depth[w] == depth[u] + 1) { // Add the arc uw
				f[i].v = w;
				f[i].next = aux[u];
				aux[u] = &(f[i++]);
			}
		}
	}

	// DFS for finding disjoint augmenting paths
	for (u = 1, tail = 0; u <= N; u++) {
		if (depth[u] != 0) continue;
		w = DFS_bipartite_matching(aux, par, u);
		if (w > 0) q[tail++] = w; // An augmenting path from u to w has been found
	}
	
	// Augmentation
	for (head = 0; head < tail; head++) {
		for (w = q[head], u = par[w]; u > 0; w = par[u], u = par[w]) {
			mate[u] = w;
			mate[w] = u;
		}
	}
	return tail;
}

int bipartite_matching(int N, char color[], edge* adj[], int mate[])
{
	int i, u, dif, ans = 0;
	edge *p;
	for (u = 1; u <= N; u++) mate[u] = 0; // Initialization
	do { // Augmentation
		dif = bipartite_matching_augmentation(N, color, adj, mate);
		ans += dif;
	} while (dif != 0);
	return ans;
}



// 2. Naive solution (O(sqrt{N + M} L^2) time)
void naive2(int N, int M, int L, int s[], int t[], char ans[])
{
	static char color[N_MAX + M_MAX + 1];
	static int i, u, w, mate[N_MAX + M_MAX + 1];
	static edge *adj[N_MAX + M_MAX + 1], e[L_MAX * 2 + 1], *p;
	for (u = 1; u <= N + M; u++) {
		adj[u] = NULL;
		color[u] = (u > N)? 1: 0;
	}
	for (i = 0; i < L; i++) {
		u = s[i+1];
		w = t[i+1] + N;
		e[i*2].v = w;
		e[i*2].next = adj[u];
		adj[u] = &(e[i*2]);
		e[i*2+1].v = u;
		e[i*2+1].next = adj[w];
		adj[w] = &(e[i*2+1]);
	}
	
	int j, x, y, mu = bipartite_matching(N + M, color, adj, mate);
	for (j = 0; j < L; j++) {
		x = s[j+1];
		y = t[j+1] + N;
		for (u = 1; u <= N + M - 2; u++) {
			adj[u] = NULL;
			color[u] = (u > N - 1)? 1: 0;
		}
		for (i = 0; i < L; i++) {
			u = s[i+1];
			w = t[i+1] + N;
			if (u == x || y == w) continue;
			if (u > x) u--;
			if (w > y) w -= 2;
			else w--;
			e[i*2].v = w;
			e[i*2].next = adj[u];
			adj[u] = &(e[i*2]);
			e[i*2+1].v = u;
			e[i*2+1].next = adj[w];
			adj[w] = &(e[i*2+1]);
		}
		if (mu == bipartite_matching(N + M - 2, color, adj, mate) + 1) ans[j+1] = 1;
		else ans[j+1] = 0;
	}
}



int main()
{
	char ans[L_MAX + 1];
	int i, N, M, L, s[L_MAX + 1], t[L_MAX + 1];
	scanf("%d %d %d", &N, &M, &L);
	for (i = 1; i <= L; i++) scanf("%d %d", &(s[i]), &(t[i]));
	naive2(N, M, L, s, t, ans);
	for (i = 1; i <= L; i++) {
		if (ans[i] == 0) printf("No\n");
		else printf("Yes\n");
	}
	fflush(stdout);
	return 0;
}
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