hdu 1690 Bus System（Floyd）

标签：floyd   最短路   

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1690

Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now.
The bus system of City X is quite strange. Unlike other city’s system, the cost of ticket is calculated based on the distance between the two stations. Here is a list which describes the relationship between the distance and the cost.



Your neighbor is a person who is a really miser. He asked you to help him to calculate the minimum cost between the two stations he listed. Can you solve this problem for him?
To simplify this problem, you can assume that all the stations are located on a straight line. We use x-coordinates to describe the stations’ positions.

2
1 2 3 4 1 3 5 7
4 2
1
2
3
4
1 4
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4

Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.

#include <cstdio>
#include <cstring>
#include <cmath>
#define INF 1e18
#define MAXN 117
__int64 dis[MAXN][MAXN];
__int64 L1, L2, L3, L4;
__int64 C1, C2, C3, C4;
__int64 f[MAXN];
int n, m;
__int64 min(__int64 a, __int64 b)
{
    return a < b ? a:b;
}
void init()
{
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            dis[i][j] = INF;
        }
    }
}
void Floyd()
{
    for(int k = 1; k <= n; k++)
    {
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                dis[i][j] = min(dis[i][j],dis[i][k]+dis[k][j]);
            }
        }
    }
}

__int64 cal(__int64 x)
{
    if(x > 0 && x <= L1)
        return C1;
    if(x > L1 && x <= L2)
        return C2;
    if(x > L2 && x <= L3)
        return C3;
    if(x > L3 && x <= L4)
        return C4;

        return INF;
}

int main()
{
    int t;
    int i, j;
    __int64 s, e;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d%I64d%I64d%I64d",&L1,&L2,&L3,&L4);
        scanf("%I64d%I64d%I64d%I64d",&C1,&C2,&C3,&C4);
        scanf("%d%d",&n,&m);
        init();
        memset(f,0,sizeof(f));
        for(i = 1; i <= n; i++)
        {
            scanf("%I64d",&f[i]);
        }
        for(i = 1; i < n; i++)
        {
            for(j = i+1; j <= n; j++)
            {
                dis[i][j] = dis[j][i] = cal(fabs(f[i]-f[j]));
            }
        }
        
        Floyd();
        
        printf("Case %d:\n",++cas);
        for(i = 1; i <= m; i++)
        {
            scanf("%I64d%I64d",&s,&e);
            if(dis[s][e] != INF)
                printf("The minimum cost between station %I64d and station %I64d is %I64d.\n",s,e,dis[s][e]);
            else
                printf("Station %I64d and station %I64d are not attainable.\n",s,e);
        }
    }
    return 0;
}

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hdu 1690 Bus System（Floyd）

标签：floyd   最短路   

原文地址：http://blog.csdn.net/u012860063/article/details/38401219