hdu 3665
来源:互联网 发布:php 微信退款接口demo 编辑:程序博客网 时间:2024/06/15 09:44
题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=3665
Seaside
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1504 Accepted Submission(s): 1087
Problem Description
XiaoY is living in a big city, there are N towns in it and some towns near the sea. All these towns are numbered from 0 to N-1 and XiaoY lives in the town numbered ’0’. There are some directed roads connecting them. It is guaranteed that you can reach any town from the town numbered ’0’, but not all towns connect to each other by roads directly, and there is no ring in this city. One day, XiaoY want to go to the seaside, he asks you to help him find out the shortest way.
Input
There are several test cases. In each cases the first line contains an integer N (0<=N<=10), indicating the number of the towns. Then followed N blocks of data, in block-i there are two integers, Mi (0<=Mi<=N-1) and Pi, then Mi lines followed. Mi means there are Mi roads beginning with the i-th town. Pi indicates whether the i-th town is near to the sea, Pi=0 means No, Pi=1 means Yes. In next Mi lines, each line contains two integers SMi and LMi, which means that the distance between the i-th town and the SMi town is LMi.
Output
Each case takes one line, print the shortest length that XiaoY reach seaside.
Sample Input
5
1 0
1 1
2 0
2 3
3 1
1 1
4 100
0 1
0 1
Sample Output
2
题意让你求从0到某一个海边村庄的最短路,用floyd求即可。
代码:
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#define maxn 0x7fffffffusing namespace std;long long d[15][15];int sea[15];int v;void floyd(){ for(int k=0;k<v;k++) { for(int i=0;i<v;i++) { for(int j=0;j<v;j++) { d[i][j]=min(d[i][j],d[i][k]+d[k][j]); } } }}int main (void){ //int r=maxn; //printf("maxn=%d\n",r); while(cin>>v){ memset(sea,0,sizeof(sea)); for(int i=0;i<v;i++) { for(int j=0;j<v;j++) { d[i][j]=maxn; } d[i][i]=0; } int a,b; int c,le; for(int i=0;i<v;i++) { scanf("%d %d",&a,&b); sea[i]=b; for(int j=0;j<a;j++) { scanf("%d %d",&c,&le); d[i][c]=le; } } floyd(); long long ans=maxn; for(int i=0;i<v;i++) { // floyd(); if(sea[i]&&d[0][i]<ans) { ans=d[0][i]; // printf("%d___\n",ans); } } printf("%lld\n",ans); } return 0;}
- hdu 3665
- hdu 3665 Seaside
- 【floyd】hdu 3665
- HDU 3665 dijkstra floyd
- HDU 3665 简单floyd
- hdu 3665:Seaside
- hdu 3665(最短路)
- hdu 3665 Seaside
- HDU--3665--Seaside
- Seaside HDU 3665 【Dijkstra】
- hdu-3665 Seaside
- HDU 3665 Seaside
- hdu 3665 seaside floyd
- HDU-3665-Seaside
- HDU-3665 Seaside
- hdu
- hdu
- HDU
- 【SSH网上商城项目实战16】Hibernate的二级缓存处理首页的热门显示
- CSS 让页面内容居中显示
- KINECT+opencv(2)基于骨骼信息对视频进行动作识别
- 一种终端应用动态适配智能终端屏幕的方法及系统
- bestcoder 2016百度之星资格赛 1001 (逆元)
- hdu 3665
- 实现悬浮窗口自动吸附到屏幕边缘
- 【连载】关系型数据库是如何工作的?(15) - 查询管理器之Hash Join
- POJ 3368 RMQ
- android cordova hybrid app总结
- HDU 1098 Ignatius's puzzle
- jzoj 2016.5.14noip模拟赛C 总结
- Solr5.5使用schema.xml
- hadoop-2.6.0 完全分布式安装