hdu 1690 Bus System

题意：求到达任意俩点之间的最小消费，也可以说求任意俩点的最短路径啦

分析：很直接的就是用ｆｌｏｙｄ算法，很好写的一个算法，最坑爹的是距离要用＿＿ｉｎｔ６４保存

当然，对每一次询问都用单源最短路算法的ＳＰＦＡ　也行，不过慢了一点

#include<iostream>
#include<algorithm>
#include<math.h>
#define MAXN 100+10
using namespace std;
int L[4],C[4],n;
int X[MAXN];
__int64 g[MAXN][MAXN];
int get_dis(int i,int j)
{
    int dis=abs(X[j]-X[i]);
    if(dis>0 && dis<=L[0])
        return C[0];
    if(dis>L[0] && dis<=L[1])
        return C[1];
    if(dis>L[1] && dis<=L[2])
        return C[2];
    if(dis>L[2] && dis<=L[3])
        return C[3];
    return -1;
}
void floyd()
{
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
            {
                if(g[i][k]==-1 || g[k][j]==-1)
                    continue;
                if(g[i][j]==-1 || g[i][k]+g[k][j]<g[i][j])
                    g[i][j]=g[i][k]+g[k][j];
            }
}
int main()
{
    int T,m,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        for(int i=0;i<4;i++)
            scanf("%d",&L[i]);
        for(int i=0;i<4;i++)
            scanf("%d",&C[i]);
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&X[i]);
        for(int i=1;i<n;i++)
        {
            g[i][i]=-1;
            for(int j=i+1;j<=n;j++)
                g[i][j]=g[j][i]=get_dis(i,j);
        }
        g[n][n]=-1;
        floyd();
        int a,b;
        printf("Case %d:\n",++cas);
        while(m--)
        {
            scanf("%d %d",&a,&b);
            if(g[a][b]!=-1)
                printf("The minimum cost between station %d and station %d is %I64d.\n",a,b,g[a][b]);
            else printf("Station %d and station %d are not attainable.\n",a,b);
        }
    }
    return 0;
}

#include<iostream>
#include<algorithm>
#include<math.h>
#define MAXN 100+10
using namespace std;
int g[MAXN][MAXN],n;
int L[4],C[4];
__int64 dist[MAXN];
int X[MAXN],Q[MAXN*MAXN*MAXN];
bool vis[MAXN];
int get_dis(int i,int j)
{
    int dis=abs(X[j]-X[i]);
    if(dis>0 && dis<=L[0])
        return C[0];
    if(dis>L[0] && dis<=L[1])
        return C[1];
    if(dis>L[1] && dis<=L[2])
        return C[2];
    if(dis>L[2] && dis<=L[3])
        return C[3];
    return -1;
}
void SPFA(int s)
{    // pri是队列头结点，end是队列尾结点   
    int i, pri, end, p;  
    memset(vis, 0, sizeof(vis));
    for(int i=0; i<MAXN*MAXN; ++i)   
        Q[i] = 0;  
    for (i=1; i<=n; i++)     
        dist[i] = -1;   
    dist[s] = 0;    
    vis[s] = 1; 
    Q[1] = s; pri = 1; end = 2; 
    while (pri < end)  
    {      
        p = Q[pri];      
        for (i=1; i<=n; ++i)    
        {            //更新dis  
            if(g[p][i]==-1 || dist[p]==-1) continue;
            if (dist[i]==-1 || dist[p]+ g[p][i]<dist[i])    
            {              
                dist[i] = dist[p]+g[p][i];      
                if (!vis[i])     //未在队列中       
                {          
                    Q[end++] = i;            
                    vis[i] = 1;     
                }        
            }    
        }       
        vis[p] = 0;   // 置出队的点为未标记       
        pri++;   
    }   
} 
int main()
{
    int T,m,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        for(int i=0;i<4;i++)
            scanf("%d",&L[i]);
        for(int i=0;i<4;i++)
            scanf("%d",&C[i]);
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&X[i]);
        for(int i=1;i<n;i++)
        {
            g[i][i]=-1;
            for(int j=i+1;j<=n;j++)
                g[i][j]=g[j][i]=get_dis(i,j);
        }
        g[n][n]=-1;
        int a,b;
        printf("Case %d:\n",++cas);
        while(m--)
        {
            scanf("%d %d",&a,&b);
            SPFA(a);
            if(dist[b]!=-1)
                printf("The minimum cost between station %d and station %d is %I64d.\n",a,b,dist[b]);
            else printf("Station %d and station %d are not attainable.\n",a,b);
        }
    }
    return 0;
}