POJ 5636 Shortest Path（floyd）

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本文介绍了一种在特定路径图中寻找最短路径的方法。该方法通过对预设的额外节点进行预处理，优化了查询效率，能够快速计算任意两点间的最短距离。

Shortest Path

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1072 Accepted Submission(s): 340

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

题目大意：给一个含有n个节点的图,任意节点i和i+1之间有权值为1的边，现在往里面加入三条边权值为1的边，求S=(∑i=1mi⋅zi)，zi为第i对点之间的最短距离。

解题思路:直接对所给的6个点求任意两点的最短路，然后计算所给点对通过这些点的最短路即可。

代码如下:

/*
time:
author:tpbluesky
answer:
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <vector>
#include <set>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdlib>
#include <sstream>
#define inf 0x3f3f3f3f
#define eps 1e-6
#define sqr(x) ((x)*(x))
using namespace std;
typedef long long ll;
const int maxn = 10;
const int mod = 1000000007;

int a[maxn], b[maxn], n, m;
int mp[maxn][maxn];

int getpos(int x){
	return lower_bound(a,a+6,x)-a;
}

int main()
{
	int T;
	cin>>T;
	while(T--){
		scanf("%d%d",&n,&m);
		for(int i = 0;i < 6;++i){
			scanf("%d",&a[i]);
			b[i] = a[i];
		}
		sort(a,a+6);
		memset(mp,0,sizeof(mp));
		for(int i = 0;i < 6;++i)
			for(int j = 0;j < 6;++j)
				mp[i][j] = abs(a[i]-a[j]);
		for(int i = 0;i < 6;i+= 2){
			int x = getpos(b[i]);
			int y = getpos(b[i+1]);
			mp[x][y] = mp[y][x] = min(mp[x][y],1);
		} 
		//floyd
		for(int k = 0;k < 6;++k)
			for(int i = 0;i < 6;++i)
				for(int j = 0;j < 6;++j)
					mp[i][j] = min(mp[i][j],mp[i][k]+mp[k][j]);
		ll ans = 0;
		int x, y;
		for(int k = 1;k <= m;++k){
			scanf("%d%d",&x,&y);
			ll res = abs(x-y);
			for(int i = 0;i < 6;++i)
				for(int j = 0;j < 6;++j)
					res = min(res,(ll)(abs(x-a[i])+abs(y-a[j])+mp[i][j]));
			ans  = ((ans+(res*k)%mod)%mod);
		}
		printf("%lld\n",ans);
	}
    return 0;
}