結果
問題 | No.1744 Selfish Spies 1 (à la Princess' Perfectionism) |
ユーザー | ygussany |
提出日時 | 2021-11-08 00:32:37 |
言語 | C (gcc 12.3.0) |
結果 |
AC
|
実行時間 | 139 ms / 5,000 ms |
コード長 | 5,181 bytes |
コンパイル時間 | 367 ms |
コンパイル使用メモリ | 34,688 KB |
実行使用メモリ | 7,564 KB |
最終ジャッジ日時 | 2024-05-07 12:18:56 |
合計ジャッジ時間 | 2,471 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1 ms
6,816 KB |
testcase_01 | AC | 1 ms
6,940 KB |
testcase_02 | AC | 1 ms
6,944 KB |
testcase_03 | AC | 1 ms
6,940 KB |
testcase_04 | AC | 2 ms
6,940 KB |
testcase_05 | AC | 1 ms
6,944 KB |
testcase_06 | AC | 1 ms
6,940 KB |
testcase_07 | AC | 1 ms
6,940 KB |
testcase_08 | AC | 1 ms
6,944 KB |
testcase_09 | AC | 1 ms
6,944 KB |
testcase_10 | AC | 1 ms
6,940 KB |
testcase_11 | AC | 1 ms
6,940 KB |
testcase_12 | AC | 1 ms
6,940 KB |
testcase_13 | AC | 1 ms
6,940 KB |
testcase_14 | AC | 2 ms
6,940 KB |
testcase_15 | AC | 2 ms
6,940 KB |
testcase_16 | AC | 2 ms
6,940 KB |
testcase_17 | AC | 3 ms
6,940 KB |
testcase_18 | AC | 3 ms
6,940 KB |
testcase_19 | AC | 1 ms
6,944 KB |
testcase_20 | AC | 1 ms
6,944 KB |
testcase_21 | AC | 1 ms
6,944 KB |
testcase_22 | AC | 1 ms
6,944 KB |
testcase_23 | AC | 1 ms
6,944 KB |
testcase_24 | AC | 8 ms
6,940 KB |
testcase_25 | AC | 2 ms
6,944 KB |
testcase_26 | AC | 2 ms
6,940 KB |
testcase_27 | AC | 4 ms
6,940 KB |
testcase_28 | AC | 76 ms
7,564 KB |
testcase_29 | AC | 4 ms
6,940 KB |
testcase_30 | AC | 4 ms
6,940 KB |
testcase_31 | AC | 4 ms
6,940 KB |
testcase_32 | AC | 4 ms
6,944 KB |
testcase_33 | AC | 75 ms
7,400 KB |
testcase_34 | AC | 62 ms
7,424 KB |
testcase_35 | AC | 139 ms
7,312 KB |
testcase_36 | AC | 137 ms
7,308 KB |
testcase_37 | AC | 138 ms
7,564 KB |
testcase_38 | AC | 123 ms
7,432 KB |
ソースコード
#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; } 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; } // 1. Solution example (O(sqrt{N + M} L) time) void solve1(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; } } if (p != NULL) break; } 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; } } if (p != NULL) break; } 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++]); } } } static int num[N_MAX + M_MAX + 1]; int n = SCC(N + M, aux, label); for (i = 1; i <= n; i++) num[i] = 0; for (u = 1; u <= N + M; u++) num[label[u]]++; for (i = 1; i <= L; i++) { u = s[i]; w = t[i] + N; if (par[u] == 0 && par[w] == 0 && label[u] == label[w] && num[label[u]] == 2) ans[i] = 0; else ans[i] = 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])); solve1(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; }