結果
問題 | No.1814 Uribo Road (Easy) |
ユーザー |
👑 |
提出日時 | 2022-01-15 20:01:26 |
言語 | C (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 24 ms / 2,000 ms |
コード長 | 2,110 bytes |
コンパイル時間 | 295 ms |
コンパイル使用メモリ | 32,000 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-11-21 23:51:16 |
合計ジャッジ時間 | 1,582 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 |
---|---|
sample | AC * 3 |
other | AC * 44 |
ソースコード
#include <stdio.h>const int bit[13] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096};typedef struct {int key, id[2];} data;typedef struct {data obj[40000000];int size;} min_heap;void push(min_heap* h, data x){int i = ++(h->size), j = i >> 1;data tmp;h->obj[i] = x;while (j > 0) {if (h->obj[i].key < h->obj[j].key) {tmp = h->obj[j];h->obj[j] = h->obj[i];h->obj[i] = tmp;i = j;j >>= 1;} else break;}}data pop(min_heap* h){int i = 1, j = 2;data output = h->obj[1], tmp;h->obj[1] = h->obj[(h->size)--];while (j <= h->size) {if (j < h->size && h->obj[j^1].key < h->obj[j].key) j ^= 1;if (h->obj[j].key < h->obj[i].key) {tmp = h->obj[j];h->obj[j] = h->obj[i];h->obj[i] = tmp;i = j;j <<= 1;} else break;}return output;}typedef struct Edge {struct Edge *next;int v, cost, flag;} edge;int main(){int i, j, N, M, K, u, w, c, R[12];edge *adj[10001] = {}, e[40001], *p;scanf("%d %d %d", &N, &M, &K);for (i = 0; i < K; i++) scanf("%d", &(R[i]));for (i = 0; i < M; i++) {scanf("%d %d %d", &u, &w, &c);e[i*2].v = w;e[i*2+1].v = u;e[i*2].cost = c;e[i*2+1].cost = c;e[i*2].flag = 0;e[i*2+1].flag = 0;for (j = 0; j < K; j++) {if (R[j] == i + 1) {e[i*2].flag = bit[j];e[i*2+1].flag = bit[j];}}e[i*2].next = adj[u];e[i*2+1].next = adj[w];adj[u] = &(e[i*2]);adj[w] = &(e[i*2+1]);}const int sup = ((1 << 30) - 1) << 1;int dist[10001][4096];data d;min_heap h;h.size = 0;for (u = 1; u <= N; u++) for (i = 0; i < bit[K]; i++) dist[u][i] = sup;dist[1][0] = 0;d.key = 0;d.id[0] = 1;d.id[1] = 0;push(&h, d);while (h.size > 0) {d = pop(&h);u = d.id[0];i = d.id[1];if (d.key != dist[u][i]) continue;for (p = adj[u]; p != NULL; p = p->next) {w = p->v;j = i | p->flag;if (dist[w][j] > dist[u][i] + p->cost) {dist[w][j] = dist[u][i] + p->cost;d.key = dist[w][j];d.id[0] = w;d.id[1] = j;push(&h, d);}}}printf("%d\n", dist[N][bit[K]-1]);fflush(stdout);return 0;}