HDOJ--2680--Choose the best route

Choose the best route

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10372    Accepted Submission(s): 3342

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：琪琪想要去拜訪她的朋友，可是这货easy晕车，所以要找一个花费时间最少的路线。

如今给你路线图。让你找出从她家附近的起点站（能够有多个）到朋友家附近的终点站（仅仅有一个）花费时间最少的路线。各个网站的编号从1到n。 

思路：最開始我是直接无脑用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)//过滤掉同样的边。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)//推断是否存在这种值。不存在的话ans数组肯定每个都是INF。 
			printf("-1
");
		else
			printf("%d
",ans[0]);
	}
	return 0;
}

再次使用SPFA过一次。这道题非常坑，哦，不，非常坑。假设你数组开小了。用G++提交会显示超时而不是RE，不明觉厉。

ac代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 0x3f3f3f3f
#define N 1010
#define M 20100
using namespace std;
int dis[N],vis[N],n,m,edgenum,head[N];
struct node{
	int from,to,cost,next;
}edge[M];//结构体是用来记录边的数量的。应该开到M； 
void init(){
	edgenum=0;
	memset(head,-1,sizeof(head));//表头应注意初始时为-1。 
}
void add(int u,int v,int cost){
	node E={u,v,cost,head[u]};
	edge[edgenum]=E;
	head[u]=edgenum++;
}
void spfa(int beg){
	queue<int>q;
	memset(dis,INF,sizeof(dis));
	memset(vis,0,sizeof(vis));
	dis[beg]=0;
	vis[beg]=1;
	q.push(beg);
	while(!q.empty()){
		int i,u=q.front();
		q.pop();
		vis[u]=0;
		for(i=head[u];i!=-1;i=edge[i].next){
			int v=edge[i].to;
			if(dis[v]>dis[u]+edge[i].cost){
				dis[v]=dis[u]+edge[i].cost;
				if(!vis[v]){
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
}
int main(){
	int beg;
	while(scanf("%d%d%d",&n,&m,&beg)!=EOF){
		init();
		while(m--){
			int a,b,c;
			scanf("%d%d%d",&a,&b,&c);
			add(b,a,c);
		}
		spfa(beg);
		int e,min=INF;
		scanf("%d",&e);
		while(e--){
			int end;
			scanf("%d",&end);
			if(min>dis[end])
				min=dis[end];
		}
		if(min==INF)
			printf("-1
");
		else
			printf("%d
",min);
	}
	return 0;
}