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): 10619    Accepted Submission(s): 3423

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
 
 
代码1：反向考虑；求一次dijkstra即可。 530MS。
 
#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3fusing namespace std;int map[1010][1010],low[1010],vis[1010],time;int n;int init(){    for(int i=1;i<=n;i++)    {        for(int j=1;j<=n;j++)        {            if(i==j)            map[i][j]=0;            else            map[i][j]=INF;        }    }}void dijkstra(int x){    int min,i,j,next;    memset(vis,0,sizeof(vis));    for(i=1;i<=n;i++)    {        low[i]=map[x][i];    }    vis[x]=1;    for(i=2;i<=n;i++)    {        min=INF;        for(j=1;j<=n;j++)        {            if(!vis[j]&&min>low[j])            {                min=low[j];                next=j;            }        }        if(min==INF)        break;        vis[next]=1;        for(j=1;j<=n;j++)        {            if(!vis[j]&&low[j]>low[next]+map[next][j])            {                low[j]=low[next]+map[next][j];            }        }    }}int main(){    int i,a,b,c,w,m,s,beg;    while(scanf("%d%d%d",&n,&m,&s)!=EOF)    {        time=INF;        init();        for(i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            if(map[b][a]>c) //反向求；两车站间的用时最少的路。             map[b][a]=c;        }        dijkstra(s);        scanf("%d",&w);        for(i=0;i<w;i++)        {            scanf("%d",&beg);            if(time>low[beg])            time=low[beg];        }        if(time==INF)        printf("-1\n");        else        printf("%d\n",time);    }    return 0;}
 
 
 
代码2：正向考虑，求一次dijkstra即可。 452MS。
 
 
#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3fusing namespace std;int map[1010][1010],low[1010],vis[1010];int n,m,s;int init(){    for(int i=0;i<=n;i++)    {        for(int j=0;j<=n;j++)        {            if(i==j)            map[i][j]=0;            else            map[i][j]=INF;        }    }}void dijkstra(int x){    int i,j,min,next;    memset(vis,0,sizeof(vis));    for(i=0;i<=n;i++)    {        low[i]=map[x][i];    }    vis[x]=1;    for(i=0;i<=n;i++)    {        min=INF;        for(j=0;j<=n;j++)        {            if(!vis[j]&&min>low[j])            {                min=low[j];                next=j;            }        }        vis[next]=1;        if(min==INF)        {            break;        }        for(j=0;j<=n;j++)        {            if(!vis[j]&&low[j]>low[next]+map[next][j])            {                low[j]=low[next]+map[next][j];            }        }    }}int main(){    int i,a,b,c,w,beg;    while(scanf("%d%d%d",&n,&m,&s)!=EOF)    {        init();        for(i=0;i<m;i++)        {            scanf("%d%d%d",&a,&b,&c);            if(map[a][b]>c)            {                map[a][b]=c;            }        }        scanf("%d",&w);        for(i=0;i<w;i++)        {            scanf("%d",&beg);            map[0][beg]=0; //设0到起点的距离为0；         }        dijkstra(0);        if(low[s]==INF)        printf("-1\n");        else        printf("%d\n",low[s]);    }    return 0;}
 

代码3:spfa
 
#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>#define MAXN 1000+10#define MAXM 20000+10#define INF 0x3f3f3fusing namespace std;struct record{    int to,next,val;}edge[MAXM];int vis[MAXN],low[MAXN],head[MAXN];int top,n,m,s;queue<int>q;void init(){    top=0;    for(int i=1;i<=n;i++)    {        head[i]=-1;        vis[i]=0;    }}void add(int a,int b,int c){    edge[top].to=b;    edge[top].val=c;    edge[top].next=head[a];    head[a]=top++;}void spfa(int x){    int i,j;    for(i=1;i<=n;i++)    {        low[i]=INF;    }    vis[x]=1;    low[x]=0;    q.push(x);    while(!q.empty())    {        int u=q.front();        q.pop();        vis[u]=0;        for(i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(low[v]>low[u]+edge[i].val)            {                low[v]=low[u]+edge[i].val;                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }    }}void get(){    int a,b,c;    while(m--)    {        scanf("%d%d%d",&a,&b,&c);        add(b,a,c);    }}void solve(){    int i,time,w,beg;    time=INF;    spfa(s);    scanf("%d",&w);    for(i=0;i<w;i++)    {        scanf("%d",&beg);        time=min(time,low[beg]);    }    if(time==INF)    printf("-1\n");    else    printf("%d\n",time);}int main(){    while(scanf("%d%d%d",&n,&m,&s)!=EOF)    {        init();        get();        solve();    }    return 0;}