題目意思：給出各個長度和花費，然後給出n,m, 接下來n行每行一個數，代表0到城市i的距離。注意這些城市是在一條直線上的。然後詢問m個城市間的最小花費。

思路：用flody求一下就行，注意開long long，INF要初始化成0x3f3f3f3f3f3f3f3f

AC程式碼：

#include <iostream>
#include <cstring>
#include <cmath>
using namespace std;
#define ll long long
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int MAXF = 4;
const int MAXN = 105;
ll L[MAXF], C[MAXF];
ll X[MAXN];
ll cost[MAXN][MAXN];

ll getC(ll dis)
{
    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 INF;
}

void floyd(int n)
{
    for (int k = 0; k < n; k++)
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                cost[i][j] = min(cost[i][j], cost[i][k] + cost[k][j]);
}

int main(int argc, const char * argv[])
{
    int T;
    cin >> T;
    int n, m;
    for (int i = 1; i <= T; i++)
    {
        cin >> L[0] >> L[1] >> L[2] >> L[3];
        cin >> C[0] >> C[1] >> C[2] >> C[3];
        cin >> n >> m;
        for (int j = 0; j < n; j++)
            scanf("%lld", X + j);
        memset(cost, 0x3f, sizeof(cost));

        for (int j = 0; j < n; j++)
            for (int k = j + 1; k < n; k++)
                cost[j][k] = cost[k][j] = getC(abs(X[j] - X[k]));
        floyd(n);
        int u, v;
        printf("Case %d:\n", i);
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d", &u, &v);
            if(cost[u - 1][v - 1] != INF)
                printf("The minimum cost between station %d and station %d is %lld.\n",
                       u, v, cost[u - 1][v - 1]);
            else  printf("Station %d and station %d are not attainable.\n", u, v);
        }
    }

    return 0;
}