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 20:32:29
言語 C
(gcc 13.3.0)
結果
TLE  
(最新)
AC  
(最初)
実行時間 -
コード長 5,056 bytes
コンパイル時間 409 ms
コンパイル使用メモリ 34,296 KB
実行使用メモリ 30,472 KB
最終ジャッジ日時 2024-11-30 05:00:31
合計ジャッジ時間 61,477 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
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ファイルパターン 結果
other AC * 52 TLE * 7
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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 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;
}
void chmin(int* a, int b)
{
    if (*a > b) *a = b;
}
int DFS_SCC(edge* adj[], int label[], int ord[], int low[], int s[], int* head, int u)
{
    s[(*head)++] = u; // Add u to the stack (which maintains the vertices already found but not determined)
    ord[u] = ord[0]++;
    low[u] = ord[u];
    
    int w;
    edge *p;
    for (p = adj[u]; p != NULL; p = p->next) {
        w = p->v;
        if (ord[w] == 0) chmin(&(low[u]), DFS_SCC(adj, label, ord, low, s, head, w)); // w has been found
        else if (ord[w] <= N_MAX + M_MAX) chmin(&(low[u]), ord[w]); // w is already found but not determined
    }
    
    if (low[u] == ord[u]) { // A new SCC containing u has been determined
        while (s[--(*head)] != u) {
            label[s[*head]] = label[0];
            ord[s[*head]] = N_MAX + M_MAX + 1;
        }
        label[u] = label[0]++;
        ord[u] = N_MAX + M_MAX + 1;
    }
    return low[u];
}
int SCC(int N, edge* adj[], int label[])
{
    int u, w, head;
    static int ord[N_MAX + M_MAX + 1], low[N_MAX + M_MAX + 1], s[N_MAX + M_MAX + 1];
    for (u = 1; u <= N; u++) {
        label[u] = 0;
        ord[u] = 0;
    }
    for (u = 1, label[0] = 1, ord[0] = 1; u <= N; u++) {
        if (ord[u] != 0) continue;
        head = 0;
        DFS_SCC(adj, label, ord, low, s, &head, u);
    }
    return label[0] - 1;
}
// 2. Solution example (O(sqrt{N + M} L) time)
void solve2(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]);
    }
    bipartite_matching(N + M, color, adj, mate);
    
    static int par[N_MAX + M_MAX + 1], q[N_MAX + M_MAX + 1], head, tail;
    for (u = 1; u <= N + M; u++) par[u] = 0;
    for (u = 1, tail = 0; u <= N; u++) {
        if (mate[u] == 0) {
            par[u] = u;
            q[tail++] = u;
        }
    }
    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;
                    q[tail++] = w;
                }
            }
        } else {
            par[mate[u]] = u;
            q[tail++] = mate[u];
        }
    }
    for (u = N + 1, tail = 0; u <= N + M; u++) {
        if (mate[u] == 0) {
            par[u] = -u;
            q[tail++] = u;
        }
    }
    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;
                    q[tail++] = w;
                }
            }
        } else {
            par[mate[u]] = -u;
            q[tail++] = mate[u];
        }
    }
    
    int m = 0;
    static int label[N_MAX + M_MAX + 1];
    static edge *aux[N_MAX + M_MAX + 1], f[L_MAX + N_MAX + 1];
    for (u = 1; u <= N; u++) {
        aux[u] = NULL;
        if (par[u] != 0) continue;
        for (p = adj[u]; p != NULL; p = p->next) {
            w = p->v;
            if (par[w] != 0) continue;
            f[m].v = w;
            f[m].next = aux[u];
            aux[u] = &(f[m++]);
        }
    }
    for (u = N + 1; u <= N + M; u++) {
        aux[u] = NULL;
        if (par[u] != 0) continue;
        for (p = adj[u]; p != NULL; p = p->next) {
            w = p->v;
            if (par[w] == 0 && w == mate[u]) {
                f[m].v = w;
                f[m].next = aux[u];
                aux[u] = &(f[m++]);
            }
        }
    }
    SCC(N + M, aux, label);
    
    for (i = 1; i <= L; i++) {
        u = s[i];
        w = t[i] + N;
        if (par[u] > 0 && par[w] > 0) ans[i] = 1;
        else if (par[u] < 0 && par[w] < 0) ans[i] = 1;
        else if (label[u] == label[w]) ans[i] = 1;
        else ans[i] = 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]));
    solve2(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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