Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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.

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

题意：琪琪想要去拜访她的朋友，但是这货容易晕车，所以要找一个花费时间最少的路线。现在给你路线图，让你找出从她家附近的起点站（可以有多个）到朋友家附近的终点站（只有一个）花费时间最少的路线。各个站点的编号从1到n。

思路：最开始我是直接无脑用DIJ算法做的结果丝毫不意外（TLE）了，FUCK。看了看别人的思路才造应该把终点当起点反向构图，注意，这是个单向图。

AC代码：

#include<stdio.h>
#include<string.h>
#define INF 0x3f3f3f3f
#include<algorithm>
using namespace std;
int vis[1010],map[1010][1010],dis[1010],n,ans[1010],num,beg;
void init(){
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++){
			if(i==j)
				map[i][j]=map[j][i]=0;
			else
				map[i][j]=map[j][i]=INF;
		}
}
void dijkstra(){
	int k=0,flag=0,i;
		memset(vis,0,sizeof(vis));
		for(i=1;i<=n;i++)
			dis[i]=map[beg][i];
		vis[beg]=1;
		for(i=1;i<=n;i++){
			int j,key,temp=INF;
			for(int j=1;j<=n;j++)
				if(!vis[j]&&temp>dis[j])
					temp=dis[key=j];
			if(temp==INF){
				break;
			}
			vis[key]=1;
			for(int j=1;j<=n;j++)
				if(!vis[j]&&dis[j]>dis[key]+map[key][j])
					dis[j]=dis[key]+map[key][j];
		}
}
int main(){
	int m;
	while(scanf("%d%d%d",&n,&m,&beg)!=EOF){
		int i;
		memset(ans,INF,sizeof(INF));
		init();//注意要初始化。
		for(i=1;i<=m;i++){
			int a,b,cost;
			scanf("%d%d%d",&a,&b,&cost);
			if(map[b][a]>cost)//过滤掉相同的边。反向构图。
				map[b][a]=cost;
		}
		scanf("%d",&num);
		dijkstra();//直接找到各个终点的位置。
		for(i=0;i<num;i++){
			int end;
			scanf("%d",&end);
			ans[i]=dis[end];
		}
		sort(ans,ans+num);
		if(ans[0]==INF)//判断是否存在这样的值，不存在的话ans数组肯定每一个都是INF。
			printf("-1\n");
		else
			printf("%d\n",ans[0]);
	}
	return 0;
}

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