Elevators UVALive

来源：互联网发布：简历管理系统知乎编辑：程序博客网时间：2024/05/20 01:46

题目链接：https://vjudge.net/problem/UVALive-7015

题意：n个电梯，每个电梯只能在特定的楼层停，若不换乘电梯，则从楼层x到楼层y所花费时间为abs(x-y)，在某一楼层换乘电梯不消耗时间。求从当前所在楼层到目的楼层所需最小时间。

思路：输入中给的同一电梯相邻楼层建边，floyd求一次最短路，然后直接输出即可。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int  maxn = 2000 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const int inf = 0x3f3f3f3f;const int MOD = 1000;int n, m, k;int dis[maxn][maxn];int num[maxn];int maxnnum;int floyd(){    for(int k = 0; k <= 150; ++k){        for(int i = 0; i <= 150; ++i){            for(int j = 0; j <= 150; ++j){                dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);            }        }    }}int main(){    int a, b;    while(~scanf("%d%d%d", &n, &a, &b)){        if(!n && !a && !b) break;        for(int i = 0; i <= 150; ++i){            dis[i][i] = 0;            for(int j = i + 1; j <= 150; ++j){                dis[i][j] = dis[j][i] = inf;            }        }        memset(num, 0, sizeof num);        int x, u, v;        for(int i = 0; i < n; ++i){            scanf("%d", &x);            scanf("%d", &num[0]);            for(int j = 1; j < x; ++j){                scanf("%d", &num[j]);                dis[num[j - 1]][num[j]] = dis[num[j]][num[j - 1]] = abs(num[j] - num[j - 1]);            }        }        floyd();        printf("%d\n", dis[a][b]);    }    return 0;}

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