hdu5365Shortest Path (floyd)

最新推荐文章于 2021-05-23 08:20:46 发布

Kirito_Acmer

最新推荐文章于 2021-05-23 08:20:46 发布

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分类专栏：最短路文章标签： floyd 最短路

本文链接：https://blog.youkuaiyun.com/Kirito_Acmer/article/details/50858488

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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<n). To make the graph more interesting, someone adds three more edges to the graph. The length of each new edge is

1.

You are given the graph and several queries about the shortest path between some pairs of vertices.

Input

There are multiple test cases. The first line of input contains an integer

T, indicating the number of test cases. For each test case:

The first line contains two integer

n and

(1≤n,m≤105) -- the number of vertices and the number of queries. The next line contains 6 integers

a1,b1,a2,b2,a3,b3

(1≤a1,a2,a3,b1,b2,b3≤n), separated by a space, denoting the new added three edges are

(a1,b1),

(a2,b2),

(a3,b3).

In the next

m lines, each contains two integers

si and

(1≤si,ti≤n), denoting a query.

The sum of values of

m in all test cases doesn't exceed

106.

Output

For each test cases, output an integer

S=(∑i=1mi⋅zi) mod (109+7), where

zi is the answer for

i-th query.

Sample Input

1
10 2
2 4 5 7 8 10
1 5
3 1

Sample Output

7
题意：给你一条n个点组成的链，相邻两点的距离为1，再给你三条边，这三条边的端点都是链上的点，且每一条的距离为1。有m个询问，问你对于每两个点，从一个端点到另一个端点的最近距离是多少。
思路：可以先初始化3条边中6个点两两之间的最短距离，这个可以用floyd做，那么对于每一个询问，两个点x1,x2的最短距离为不经过任何点，或者经过3条边中的某些边，又因为我们已经初始化出3条边中任意两个点的最短距离，所以我们只要枚举a,b,即x1到a,a到b，再b到x2的最近距离。这一题floyd初始化时关键，如果每次直接8个点floyd时间复杂度就爆了。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 1000000007
#define pi acos(-1.0)
#define MOD 1000000007
int dist[10][10];
void floyd()
{
    int i,j,k;
    for(k=1;k<=6;k++){
        for(i=1;i<=6;i++){
            for(j=1;j<=6;j++){
                if(dist[i][j]>dist[i][k]+dist[k][j]){
                    dist[i][j]=dist[i][k]+dist[k][j];
                }

            }

        }

    }
}



int main()
{
    int n,m,i,j,T,k;
    int x[10];
    int a1,b1,a2,b2,a3,b3;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        scanf("%d%d%d%d%d%d",&x[1],&x[2],&x[3],&x[4],&x[5],&x[6]);
        for(i=1;i<=6;i++){
            for(j=1;j<=6;j++){
                dist[i][j]=abs(x[i]-x[j]);
            }
        }
        dist[1][2]=dist[2][1]=min(dist[1][2],1);
        dist[3][4]=dist[4][3]=min(dist[3][4],1);
        dist[5][6]=dist[6][5]=min(dist[5][6],1);
        floyd();


        ll sum=0;
        for(k=1;k<=m;k++){
            scanf("%d%d",&x[7],&x[8]);
            int ans=abs(x[7]-x[8]);
            for(i=1;i<=6;i++){
                for(j=1;j<=6;j++){
                    ans=min(ans,abs(x[7]-x[i] )+abs(x[8]-x[j])+dist[i][j] );
                    ans=min(ans,abs(x[7]-x[j] )+abs(x[8]-x[i])+dist[i][j] );

                }
            }



            sum=(sum+(ll)ans*(ll)k)%MOD;
            //printf("%d\n",floyd());


        }
        printf("%lld\n",sum);


    }
    return 0;


}