HDU 2680 Choose the best route(反向建图最短路）
Choose the best route
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15445 Accepted Submission(s): 4999
Problem Description
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.
Input
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.
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.
Output
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”.
Sample Input
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
Sample Output
1
-1
Author
dandelion
Source
Recommend
分析：题目中给的起点有很多个，终点有一个 ，那么我们可以反向建图，找终点到各个起点的最短距离
代码如下：
#include <iostream> #include <string.h> #include <stdio.h> #include <map> #include <queue> using namespace std; typedef long long LL; #define INF 0x3f3f3f3f int dis[1100]; int cost[1100][1100]; int vis[1100]; int n,m,s,x,y,val; void djs(int s) { memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) dis[i]=INF; dis[s]=0; while(1) { int v=-1; for(int i=1;i<=n;i++) if(!vis[i]&&(v==-1||dis[i]<dis[v])) v=i; if(v==-1)break; vis[v]=1; for(int i=1;i<=n;i++) dis[i]=min(dis[i],dis[v]+cost[v][i]); } } int main() { int h,p,minn; while(scanf("%d%d%d",&n,&m,&s)!=EOF) { minn=INF; for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cost[i][j]=INF; while(m--) { scanf("%d%d%d",&x,&y,&val); cost[y][x]=min(cost[y][x],val); } djs(s); scanf("%d",&h); for(int i=0;i<h;i++) { scanf("%d",&p); minn=min(minn,dis[p]); } if(minn==INF)puts("-1"); else printf("%d\n",minn); } return 0; }
