hdu 1690floyd最短路径

最新推荐文章于 2017-08-24 16:30:04 发布

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本文详细介绍了如何使用Floyd算法解决HDU 1690问题，该问题涉及求解多个站点之间的最短路径。通过分析问题背景、引入算法原理，并结合实例代码进行深入探讨，旨在帮助读者理解并掌握此类问题的求解方法。

<span style="font-family: Arial, Helvetica, sans-serif;">#include<iostream></span>

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
long long INF = 1e18;
long long L1,L2,L3,L4,C1,C2,C3,C4;
long long D[110][110];
long long f[110];
long long ABS(long long x){return x>0?x:-x;}
long long cost(long long dis){
    if(dis>0&&dis<=L1)return C1;
    if(dis<=L2)return C2;
    if(dis<=L3)return C3;
    if(dis<=L4)return C4;
    return INF;
}
void floyd(){
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(D[i][j]>D[i][k]+D[k][j])
                    D[i][j]=D[i][k]+D[k][j];
}
int main(){
//    freopen("1690.txt","r",stdin);
    int k=1,t;
    scanf("%d",&t);
    while(k<=t){
        int m,a,b;
        cout<<"Case "<<k++<<":"<<endl;
        cin>>L1>>L2>>L3>>L4;
        cin>>C1>>C2>>C3>>C4;
        cin>>n>>m;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                D[i][j]=INF;
        for(int i=1;i<=n;i++)
            cin>>f[i];
        for(int i=1;i<n;i++)
            for(int j=i+1;j<=n;j++)
                D[i][j]=D[j][i]=cost(ABS(f[i]-f[j]));
        floyd();
        for(int i=1;i<=m;i++){
            cin>>a>>b;
            if(D[a][b]!=INF)
            cout<<"The minimum cost between station "<<a<<" and station "<<b<<" is "<<D[a][b]<<"."<<endl;
            else
            cout <<"Station "<< a <<" and station "<< b <<" are not attainable."<<endl;
        }
    }
    return 0;
}

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