HDU 5636 Shortest Path(Floyed，枚举)

Shortest Path

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1146 Accepted Submission(s): 358

Problem Description
There is a path graph G=(V,E) with n vertices. Vertices are numbered from 1 to n and there is an edge with unit length between i and i+1 (1≤i

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
#define mod 1000000007
typedef long long int  LL;
int dp[10][10];
int n,m;
long long int s;
int main()
{
    int t;
    scanf("%d",&t);
    int a[10];
    int x,y;
    while(t--)
    {
        scanf("%d%d",&n,&m);
        scanf("%d%d%d%d%d%d",&a[1],&a[2],&a[3],&a[4],&a[5],&a[6]);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=6;i++)
        {
            for(int j=1;j<=6;j++)
            {
               dp[i][j]=abs(a[j]-a[i]);
            }
        }
        dp[1][2]=1;dp[3][4]=1;dp[5][6]=1;
        dp[2][1]=1;dp[4][3]=1;dp[6][5]=1;
        for(int k=1;k<=6;k++)
        {
            for(int i=1;i<=6;i++)
            {
                for(int j=1;j<=6;j++)
                {
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]);
                }
            }
        }
        int res=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            int ans=abs(y-x);
            for(int i=1;i<=6;i++)
            {
                for(int j=1;j<=6;j++)
                {
                    ans=min(ans,abs(x-a[i])+abs(y-a[j])+dp[i][j]);

                }
            }
// int num=i*ans%mod;
// res+=num;
// res%=mod;
            (res+=(LL)i*ans%mod)%=mod;

        }
        printf("%d\n",res);


    }
    return 0;
}