HDU-2680-Choose the best route

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2680

Choose the best route

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 51 2 21 5 31 3 42 4 72 5 62 3 53 5 14 5 122 34 3 41 2 31 3 42 3 211

 

Sample Output

1-1

 

dijkstra算法：输入n个车站，m条路线，t为终点，m行里分别是两个站点及他们之间的距离，输入st个可能的出发点。注意是单向边，且两个站点之间可能有多条路线，要选最短的那条。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int inf=9999999;int e[1111][1111],dis[1111],visit[1111];int n,m;void init(){    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            if(i==j)            e[i][j]=0;            else            e[i][j]=inf;        }    }}void dijkstra(int s){    int i,j,u,v,minn;    for(i=1;i<=n;i++)    {        dis[i]=e[s][i];    }    memset(visit,0,sizeof(visit));    visit[s]=1;    for(i=1;i<n;i++)    {        minn=inf;        for(j=1;j<=n;j++)        {            if(!visit[j]&&dis[j]<minn)            {                minn=dis[j];                u=j;            }        }        if(minn==inf)        break;        visit[u]=1;        for(v=1;v<=n;v++)        {            if(!visit[v]&&dis[v]>dis[u]+e[u][v])            {                dis[v]=dis[u]+e[u][v];            }        }    }}int main(){    int i,st,t,u,v,w;    int s[1111];    while(~scanf("%d%d%d",&n,&m,&t))    {        init();        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&w);            e[v][u]=min(e[v][u],w);//构建反向图         }        int flag=1;            scanf("%d",&st);        for(i=1;i<=st;i++)        {            scanf("%d",&s[i]);            }        dijkstra(t);//反向，将终点看作起点。         int ans=inf;        for(i=1;i<=st;i++)        {            if(dis[s[i]]<ans)            ans=dis[s[i]];         }        if(ans==inf)        cout<<"-1"<<endl;        else        cout<<ans<<endl;    }    return 0;}

或者 像一个人的旅行那道题一样，把起点定为0,0到题目给出的起点的距离为0.

#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int inf=9999999;int e[1111][1111],dis[1111],visit[1111];int n,m;void init(){    for(int i=0;i<=n;i++)    {        for(int j=0;j<=n;j++)        {            if(i==j)            e[i][j]=0;            else            e[i][j]=inf;        }    }}void dijkstra(){    int i,j,u,v,minn;    for(i=0;i<=n;i++)    {        dis[i]=e[0][i];    }    memset(visit,0,sizeof(visit));    visit[0]=1;    for(i=0;i<n;i++)    {        minn=inf;        for(j=0;j<=n;j++)        {            if(!visit[j]&&dis[j]<minn)            {                minn=dis[j];                u=j;            }        }        if(minn==inf)        break;        visit[u]=1;        for(v=0;v<=n;v++)        {            if(!visit[v]&&dis[v]>dis[u]+e[u][v])            {                dis[v]=dis[u]+e[u][v];            }        }    }}int main(){    int i,st,t,u,v,w;    int s[1111];    while(~scanf("%d%d%d",&n,&m,&t))    {        init();        for(i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&w);            e[u][v]=min(e[u][v],w);         }        int flag=1;            scanf("%d",&st);        for(i=0;i<st;i++)        {            scanf("%d",&s[i]); e[0][s[i]]=0;           }        dijkstra();         if(dis[t]==inf)        cout<<"-1"<<endl;        else        cout<<dis[t]<<endl;    }    return 0;}