uva 10801 - Lift Hopping(最短路Dijkstra)

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题目链接：10801 - Lift Hopping

题目大意：有一栋100层的大楼（标号为0~99），里面有n个电梯（不超过5个），以及要到达的层数（aid)，然后是每个电梯走一层所需的时间，再n行就是对应每个电梯可以到达的层数，数量不定。然后每装换一次电梯需要等待60秒，问，最快能多快到达目标层数。

解题思路：这题有点坑啊，一开始是用邻接表+Dijkstra，可是忘记考虑aid = 0的情况，这种情况下所得到的答案不需要减掉60（为了方便计算，每次到达一个楼层时间就+60），可是我没发现这个问题，就换成用邻接矩阵去做，后来发现了aid = 0的情况，可是用邻接矩阵需要注意的是更新g[a][b]的为最小值，即当两个电梯都能满足从a层到b层的时候，要选取小的值做g[a][b]的值。

1.邻接矩阵

#include <stdio.h>#include <string.h>#include <stdlib.h>const int N = 105;const int INF = 0x3f3f3f3f;int n, aid, cnt, time[N], d[N];int g[N][N];void add(int a, int b, int s) {int dis = abs(b - a) * s;if (g[a][b] > dis)g[a][b] = g[b][a] = dis;}void init() {for (int i = 0; i < n; i++)scanf("%d", &time[i]);memset(d, 0x3f, sizeof(d));memset(g, 0x3f, sizeof(g));d[0] = 0;for (int i = 0; i < n; i++) {char ch = '\0';int num[N];for (int j = 0; ch != '\n'; j++) {scanf("%d%c", &num[j], &ch);for (int k = 0; k < j; k++)add(num[j], num[k], time[i]);}}}void solve() {int vis[N];memset(vis, 0, sizeof(vis));for (int i = 0; i < 99; i++) {int x, m = INF, flag = 0;for (int j = 0; j < 100; j++)if (!vis[j] && d[j] < m) { m = d[j], x = j, flag = 1; }if (!flag) return;vis[x] = 1;for (int j = 0; j < 100; j++) {if (!vis[j] && d[j] > d[x] + g[x][j] + 60)d[j] = d[x] + g[x][j] + 60;}}}int main () {while (scanf("%d%d", &n, &aid) == 2) {init();solve();if (aid == 0) printf("0\n");else if (d[aid] != INF) printf("%d\n", d[aid] - 60);else printf("IMPOSSIBLE\n");}return 0;}

2.邻接表

#include <stdio.h>#include <string.h>#include <stdlib.h>const int N = 105;const int M = 500005;const int INF = 1 << 30;int n, aid, cnt, time[N], d[N];int first[M], next[M], u[M], v[M], w[M];void add(int a, int b, int s) {u[cnt] = a;v[cnt] = b;w[cnt] = abs(b - a) * s;next[cnt] = first[u[cnt]];first[u[cnt]] = cnt;cnt++;}void init() {for (int i = 0; i < n; i++)scanf("%d", &time[i]);for (int i = 0; i < 100; i++) {first[i] = -1;   d[i] = (i == 0 ? 0 : INF);}for (int i = 0; i < n; i++) {char ch = '\0';int num[N];for (int j = 0; ch != '\n'; j++) {scanf("%d%c", &num[j], &ch);for (int k = 0; k < j; k++)add(num[j], num[k], time[i]), add(num[k], num[j], time[i]);}}}void solve() {int vis[N];memset(vis, 0, sizeof(vis));for (int i = 0; i < 100; i++) {int x, m = INF;for (int j = 0; j < 100; j++)if (!vis[j] && d[j] < m) { m = d[j], x = j; }vis[x] = 1;for (int j = first[x]; j != -1; j = next[j]) {if (d[x] < INF && d[v[j]] > d[x] + w[j] + 60)d[v[j]] = d[x] + w[j] + 60;}}}int main () {while (scanf("%d%d", &n, &aid) == 2) {init();solve();if (aid == 0) printf("0\n");else if (d[aid] != INF) printf("%d\n", d[aid] - 60);else printf("IMPOSSIBLE\n");}return 0;}