题意:

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

给定分段函数,表示两点之间距离为一定时的花费,给定t个case,n个点,m个询问,问需要花费金额最小值

题解:

#include<bits/stdc++.h>
#define INF 0x7fffffffffffffff
#define ll long long
using namespace std;
int n,m;
int t;
int l[10],c[10];
long long dis[10000][10000];
void init()
{
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            dis[i][j]=dis[j][i]=(i==j?0:INF);
        }
    }
}
void floyed()
{
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(dis[i][k]!=INF&&dis[k][j]!=INF&&dis[i][j]>dis[i][k]+dis[k][j]){
                    dis[i][j]=dis[i][k]+dis[k][j];
                }
            }
        }
    }
}
int main()
{
    cin>>t;
    for(int k=1;k<=t;k++){
        cin>>l[1]>>l[2]>>l[3]>>l[4]>>c[1]>>c[2]>>c[3]>>c[4];
        cin>>n>>m;
        int x[1000];
        for(int i=1;i<=n;i++)
        {
            cin>>x[i];
        }
        init();
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                if(abs(x[i]-x[j])<=l[1]){
                    dis[i][j]=dis[j][i]=c[1];
                }
                else if(abs(x[i]-x[j])<=l[2]){
                    dis[i][j]=dis[j][i]=c[2];
                }else if(abs(x[i]-x[j])<=l[3]){
                    dis[i][j]=dis[j][i]=c[3];
                }else if(abs(x[i]-x[j])<=l[4]){
                    dis[i][j]=dis[j][i]=c[4];
                }
                else {
                    dis[i][j]=dis[j][i]=INF;
                }
            }
        }
        floyed();
        cout<<"Case "<<k<<":"<<endl;
        for(int i=1;i<=m;i++){
            int ax,ay;
            cin>>ax>>ay;
            if(dis[ax][ay]!=INF){
                cout<<"The minimum cost between station "<<ax<<" and station "<<ay<<" is "<<dis[ax][ay]<<"."<<endl;
            }else{
                cout<<"Station "<<ax<<" and station "<<ay<<" are not attainable."<<endl;
            }
        }
    }
}