結果

問題 No.1745 Selfish Spies 2 (à la Princess' Perfectionism)
ユーザー 👑 ygussany
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 #
プレゼンテーションモードにする

#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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