結果

問題 No.1744 Selfish Spies 1 (à la Princess' Perfectionism)
ユーザー 👑 ygussany
提出日時 2021-11-07 19:54:27
言語 C
(gcc 13.3.0)
結果
AC  
実行時間 515 ms / 5,000 ms
コード長 3,183 bytes
コンパイル時間 1,134 ms
コンパイル使用メモリ 32,256 KB
実行使用メモリ 5,888 KB
最終ジャッジ日時 2024-11-30 04:56:19
合計ジャッジ時間 4,014 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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ファイルパターン 結果
other AC * 39
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ソースコード

diff #
プレゼンテーションモードにする

#include <stdio.h>
#include <stdlib.h>
#define N_MAX 500
#define M_MAX 500
#define L_MAX 100000
typedef struct Edge {
    struct Edge *next;
    int v;
} edge;
int bipartite_matching_augmentation_naive(int N, char color[], edge* adj[], int mate[])
{
    static int i, u, w, par[N_MAX + M_MAX + 1], q[N_MAX + M_MAX + 1], head, tail;
    edge *p;
    for (u = 1, tail = 0; u <= N; u++) {
        if (color[u] == 0 && mate[u] == 0) {
            par[u] = u;
            q[tail++] = u;
        } else par[u] = 0;
    }
    for (head = 0; head < tail; head++) {
        u = q[head];
        if (color[u] == 0) {
            for (p = adj[u]; p != NULL; p = p->next) {
                w = p->v;
                if (par[w] == 0) {
                    par[w] = u;
                    if (mate[w] == 0) break;
                    q[tail++] = w;
                }
            }
            if (p != NULL) break;
        } else {
            par[mate[u]] = u;
            q[tail++] = mate[u];
        }
    }
    if (head == tail) return 0;
    
    // Augmentation
    for (u = par[w]; u != w; w = par[u], u = par[w]) {
        mate[u] = w;
        mate[w] = u;
    }
    return 1;
}
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_naive(N, color, adj, mate);
        ans += dif;
    } while (dif != 0);
    return ans;
}
// 1. Relatively Naive solution (O((N + M) L) time)
void rel_naive1(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]);
    }
    
    static int tmp_mate[N_MAX + M_MAX + 1];
    int j = 0, x, mu = bipartite_matching(N + M, color, adj, mate);
    for (i = 0; i < L; i++) {
        u = s[i+1];
        w = t[i+1] + N;
        if (mate[u] != w) {
            ans[i+1] = 1;
            continue;
        }
        for (p = adj[u]; p != NULL; p = p->next) if (mate[p->v] == 0) break;
        if (p != NULL) {
            ans[i+1] = 1;
            continue;
        }
        for (p = adj[w]; p != NULL; p = p->next) if (mate[p->v] == 0) break;
        if (p != NULL) {
            ans[i+1] = 1;
            continue;
        }
        
        for (x = 1; x <= N + M; x++) tmp_mate[x] = mate[x];
        tmp_mate[u] = 0;
        tmp_mate[w] = 0;
        if (adj[u]->v == w) adj[u] = adj[u]->next;
        else {
            for (p = adj[u]; p->next->v != w; p = p->next);
            p->next = p->next->next;
        }
        if (adj[w]->v == u) adj[w] = adj[w]->next;
        else {
            for (p = adj[w]; p->next->v != u; p = p->next);
            p->next = p->next->next;
        }
        if (bipartite_matching_augmentation_naive(N + M, color, adj, tmp_mate) == 0) ans[i+1] = 0;
        else ans[i+1] = 1;
        e[i*2].next = adj[u];
        adj[u] = &(e[i*2]);
        e[i*2+1].next = adj[w];
        adj[w] = &(e[i*2+1]);
    }
}
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]));
    rel_naive1(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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