HDU-5636(Shortest Path)（floyd最短路徑）

HDU-5636(Shortest Path)（floyd最短路徑）

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1393    Accepted Submission(s): 440

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 is1.

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

The first line contains two integer n and m(1≤n,m≤105) -- the number of vertices and the number of queries. The next line contains 6 integersa1,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 ti(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

 

注：由於此題所給的點比較多，直接用最短路徑肯定會超時。思考很久無果，於是看了別人的題解，剛看時，毫無頭緒，硬着頭皮看了很久纔看懂。

特殊處理+floyd。

可以把基本情況看做：h  x[i]  x[j]   l   ，其中h與l是待求的兩點，x[i]、x[j]，是6個點中的任意兩個點。在求待求兩點之間的最短路徑時，可以把整條路徑看做3段。 （1）h  到 x[i]  （2）x[i]到 x[j]   (3)  x[j] 到 l  。通過遍歷所有情況來獲取最短的路徑。因爲這六個點之間因爲3座橋，而使路徑變短。所以在從h到i的路徑中，有橋，則可以使路徑變短。

當情況爲：h  l  x[i]  x[j]或x[i]  x[j]  l   h時，最短路徑s，則爲：s=abs(h-l);

for(i=1;i<=6;i++)
for(j=1;j<=6;j++)
{    
   ans=min(ans,abs(h-x[i])+m1[i][j]+abs(l-x[j])) ; //h與l,x[i]與x[j]的前後關係不能確定，所以需要把h與l的位置互換來確定最小值 
   ans=min(ans,abs(l-x[i])+m1[i][j]+abs(h-x[j])) ;
}

My  solution：

/*2016.3.17*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; 
#define mod  1000000007
long  long m1[7][7];
long long  ans;
int x[7],n,m;
void floyd()//更新6個點之間的最短距離 
{
	int i,j,k;
    for(k=1;k<=6;k++)
	for(i=1;i<=6;i++)
	for(j=1;j<=6;j++)
	{
		if(m1[i][j]>(m1[i][k]+m1[k][j]))
		m1[i][j]=m1[i][k]+m1[k][j];
	}	
	return ;
}
int main()
{
	int i,j,k,h,l,p,t;
	long long   sum,q;
	scanf("%d",&t);
	while(t--)
	{
		q=1;
		sum=0;
		scanf("%d%d",&n,&m);
		for(i=1;i<=6;i++)
		{
			scanf("%d",&x[i]);
		}
		for(i=1;i<=6;i++)
		{
			for(j=1;j<=6;j++)
			{
				m1[i][j]=m1[j][i]=abs(x[i]-x[j]);//暫存6個點之間的距離 
			}
		}
		m1[1][2]=m1[2][1]=q;//把3座橋對應的頂點之間的距離填充到最短路徑上 
		m1[3][4]=m1[4][3]=q;//無向圖 
		m1[5][6]=m1[6][5]=q;
		floyd();//更新6個點之間的最短路 
		for(k=1;k<=m;k++)
		{
			scanf("%d%d",&h,&l);
			ans=abs(h-l);//待測路徑的初始值 
			for(i=1;i<=6;i++)
			for(j=1;j<=6;j++)
			{      
				ans=min(ans,abs(h-x[i])+m1[i][j]+abs(l-x[j])) ; //h與l，x[i]與x[j]的前後關係不能確定，所以需要把h與l的位置互換來確定最小值 
				ans=min(ans,abs(l-x[i])+m1[i][j]+abs(h-x[j])) ;
			}
			ans=(ans*k)%mod;
			sum=(sum+ans)%mod;
		}
		sum%=mod;
		printf("%I64d\n",sum);
	}
	
	return 0;
}