hdoj2680 Choose the best route
2015-08-20 21:46
323 查看
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10761 Accepted Submission(s): 3484[align=left]Problem Description[/align]
One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stations which are near Kiki’s
home so that she can take. You may suppose Kiki can change the bus at any station. Please find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
[align=left]Input[/align]
There are several test cases.
Each case begins with three integers n, m and s,(n<1000,m<20000,1=<s<=n) n stands for the number of bus stations in this city and m stands for the number of directed ways between bus stations .(Maybe there are several ways between two bus stations .) s stands
for the bus station that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q , t (0<t<=1000). means from station p to station q there is a way and it will costs t minutes .
Then a line with an integer w(0<w<n), means the number of stations Kiki can take at the beginning. Then follows w integers stands for these stations.
[align=left]Output[/align]
The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
[align=left]Sample Input[/align]
5 8 5 1 2 2 1 5 3 1 3 4 2 4 7 2 5 6 2 3 5 3 5 1 4 5 1 2 2 3 4 3 4 1 2 3 1 3 4 2 3 2 1 1
[align=left]Sample Output[/align]
1 -1 该题用到了最短路径万能源点,#include<queue> #include<stdio.h> #include<string.h> #define INL 0x3f3f3f3f using namespace std; int vid[10000],x[1080][1080],vist[10000]; int N,M,D; void spfa() { for(int i=0;i<=N;i++) { vist[i]=INL;vid[i]=0; } queue<int> q; vid[0]=1; q.push(0); vist[0]=0; while(!q.empty()) { int u=q.front(); q.pop(); vid[u]=0; for(int i=0;i<=N;i++) { if(vist[i]>vist[u]+x[u][i]) { vist[i]=vist[u]+x[u][i]; if(!vid[i]) { q.push(i); vid[i]=1; } } } } } int main() { while(scanf("%d%d%d",&N,&M,&D)!=EOF) { memset(x,INL,sizeof(x)); int a,b,c; for(int i=0;i<M;i++) { scanf("%d%d%d",&a,&b,&c); if(x[a][b]>c) x[a][b]=c; } int n,g,min; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&g); x[0][g]=0;//万能源点知识 } spfa(); if(vist[D]==INL) printf("-1\n"); else printf("%d\n",vist[D]); } return 0; }
相关文章推荐
- 分布式数据库一致性解决初步
- delphi开发学习五:QuickReoprt报表控件使用实例
- 从背后知道C语言程序是怎么运行的
- 演讲的注意事项
- 演讲的注意事项
- 百度.搜狐...2015产品经理面试题
- (算法)最长回文子串
- Java 的文件相关操作
- LeetCode之Word Ladder
- LeetCode257 BinaryTreePaths(打印根节点到叶子节点的左右路径) Java题解
- db2 hadr模式实现
- Android 中 Intent 的使用
- 大数相加
- 平衡二叉树研究(AVL树)
- 洛谷1108 低价购买
- TFS中删除文件和重命名文件
- 个人--与老板对话1
- TFS中怪异的版本管理问题
- 在caffe中增加和convolution相同的层
- 单实例Singleton设计模式浅析