UVA 10801 Lift Hopping

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题目链接 http://acm.hust.edu.cn/vjudge/problem/22172

题目大意：有一栋100层的大楼（标号为0~99），
给n个电梯（不超过5个），和要到的终点楼层
然后是每个电梯走一层楼所需的时间(电梯的速度)
然后给每个电梯所能到达的楼层(数量不定 1行对应1个电梯的)
然后每换一次电梯需要等待60秒，
问，最快能多快到达目标层数。

先存图坐电梯从i楼到j楼的最小距离 mp[i][j] (INF表示从i不可达到j)
我是这样想的如果当前值要更新
那么肯定是从其他电梯过来的使得dis[i]变小
然后使得dis[v] > dis[i]+mp[i][v]
因为原本存的图已经使得dis[v] = dis[i]+mp[i][v](在同一个电梯上来的)
因此判断的应该是 dis[v] > dis[i]+mp[i][v]+60

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <map>#include <set>#define pi acos(-1)#define LL long long#define INF 0x3f3f3f3fusing namespace std;typedef pair<int, int> P;const int maxn = 1e5 + 5;int mp[105][105];int elv[6];int dis[105], vis[105];vector<int>a[105];void init(){memset(mp, INF, sizeof(mp));for (int i = 0; i < 100; i++)a[i].clear();}void dijkstra(int s){int i, j, v, mni;for (i = 0; i < 100; i++){dis[i] = mp[s][i];vis[i] = 0;}dis[s] = 0;vis[s] = 1;for (i = 0; i < 100; i++){mni = INF;for (j = 0; j < 100; j++){if (!vis[j] && mni > dis[j]){mni = dis[j];v = j;}}if (mni == INF) break;vis[v] = 1;for (j = 0; j < 100; j++){if (!vis[j] && dis[j] > dis[v] + mp[v][j] + 60)dis[j] = dis[v] + mp[v][j] + 60;}}}int main(void){//freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin);int n, to, i, j, k, t, u, v, w;while (cin >> n >> to){init();for (i = 1; i <= n; i++)scanf("%d", &elv[i]);for (i = 1; i <= n; i++){while (1){scanf("%d", &t);a[i].push_back(t);char ch = getchar();if (ch == '\n')break;}}for (i = 1; i <= n; i++){for (j = 0; j < a[i].size(); j++){for (k = j+1; k < a[i].size(); k++){u = a[i][j];v = a[i][k];w = (a[i][k] - a[i][j]) * elv[i];if (mp[u][v] > w)mp[u][v] = mp[v][u] = w;}}}dijkstra(0);if (dis[to] == INF)printf("IMPOSSIBLE\n");else printf("%d\n", dis[to]);}return 0;}

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