hdu 1690 Bus System 最短路 Floyd算法。。INF一定要很大很大。。。数据类型用long long。
Bus System
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7163 Accepted Submission(s): 1853
The bus system of City X is quite strange. Unlike other city’s system, the cost of ticket is calculated based on the distance between the two stations. Here is a list which describes the relationship between the distance and the cost.
Your neighbor is a person who is a really miser. He asked you to help him to calculate the minimum cost between the two stations he listed. Can you solve this problem for him?
To simplify this problem, you can assume that all the stations are located on a straight line. We use x-coordinates to describe the stations’ positions.
Each case contains eight integers on the first line, which are L1, L2, L3, L4, C1, C2, C3, C4, each number is non-negative and not larger than 1,000,000,000. You can also assume that L1<=L2<=L3<=L4.
Two integers, n and m, are given next, representing the number of the stations and questions. Each of the next n lines contains one integer, representing the x-coordinate of the ith station. Each of the next m lines contains two integers, representing the start point and the destination.
In all of the questions, the start point will be different from the destination.
For each case,2<=N<=100,0<=M<=500, each x-coordinate is between -1,000,000,000 and 1,000,000,000, and no two x-coordinates will have the same value.
2 1 2 3 4 1 3 5 7 4 2 1 2 3 4 1 4 4 1 1 2 3 4 1 3 5 7 4 1 1 2 3 10 1 4
Case 1: The minimum cost between station 1 and station 4 is 3. The minimum cost between station 4 and station 1 is 3. Case 2: Station 1 and station 4 are not attainable.
#include <stdio.h> #include <limits.h> #define INF 110000000000 typedef long long ll ; ll graph[110][110] , p[110]; //bool visited[110] ; ll min(ll a , ll b) { return a>b?b:a ; } void floyd(int n) { for(int k = 1 ; k <= n ; ++k) { for(int i = 1 ; i <= n ; ++i) { for(int j = 1 ; j <= n ; ++j) { graph[i][j] = min(graph[i][j],graph[i][k]+graph[k][j]) ; } } } } int main() { int t , c = 1; scanf("%d",&t) ; while(t--) { ll l1,l2,l3,l4,c1,c2,c3,c4 ; scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d",&l1,&l2,&l3,&l4,&c1,&c2,&c3,&c4) ; int n , m ; scanf("%d%d",&n,&m) ; for(int i = 1 ; i <= n ; ++i) { scanf("%I64d",&p[i]) ; } for(int i = 1 ; i <= n ; ++i) { graph[i][i] = 0 ; for(int j = 1 ; j < i ; ++j) { ll dis = p[i]-p[j] ; dis = dis>0?dis:-dis ; if(dis<=l1) { graph[i][j] = graph[j][i] = c1 ; } else if(dis<=l2) { graph[i][j] = graph[j][i] = c2 ; } else if(dis<=l3) { graph[i][j] = graph[j][i] = c3 ; } else if(dis<=l4) { graph[i][j] = graph[j][i] = c4 ; } else { graph[i][j] = graph[j][i] = INF ; } } } floyd(n) ; printf("Case %d:\n",c++) ; for(int i = 0 ; i < m ; ++i) { int s , t ; scanf("%d%d",&s,&t) ; if(graph[s][t] >= INF) { printf("Station %d and station %d are not attainable.\n",s,t) ; } else { printf("The minimum cost between station %d and station %d is %I64d.\n",s,t,graph[s][t]) ; } } } return 0 ; }
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