题目描述

For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pasture.

Each trail connects two different intersections (1 ≤ I1i ≤ 1,000; 1 ≤ I2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the lengthi of each trail (1 ≤ lengthi  ≤ 1,000), the two intersections the trail connects, and they know that no two intersections are directly connected by two different trails. The trails form a structure known mathematically as a graph.

To run the relay, the N cows position themselves at various intersections (some intersections might have more than one cow). They must position themselves properly so that they can hand off the baton cow-by-cow and end up at the proper finishing place.

Write a program to help position the cows. Find the shortest path that connects the starting intersection (S) and the ending intersection (E) and traverses exactly N cow trails.

给出一张无向连通图，求S到E经过k条边的最短路。

输入输出格式

输入格式：

 

* Line 1: Four space-separated integers: N, T, S, and E

* Lines 2..T+1: Line i+1 describes trail i with three space-separated integers: lengthi , I1i , and I2i

 

输出格式：

 

* Line 1: A single integer that is the shortest distance from intersection S to intersection E that traverses exactly N cow trails.

 

输入输出样例

输入样例#1： 

2 6 6 411 4 64 4 88 4 96 6 82 6 93 8 9

输出样例#1： 

10 题解： 一句话题意：给出一个有t条边的图，求从s到e恰好经过k条边的最短路。 a:是图的邻接矩阵，f是图中任意两点直接的最短距离、 floyd算法: f[i][j]=min(f[i][j],f[i][k]+f[k][j]); 一遍floyed后： f[i][j]是i到j的最短距离：中间至少1条边（连通图），最多n-1条边

floyd算法的变形：

a:是图的邻接矩阵，a[i][j]是经过一条边的最短路径。f[i][j]k的初值为∞

f[i][j]-1=∞;

f[i][j]1=a;    //经过一条边的最短路径

f[i][j]2=min(f[i][j]2,a[i][k]+a[k][j])=a*a=a2;//经过二条边的最短路径，经过一次floyd.矩阵相乘一次,f[i][j]2初值为∞。

f[i][j]3=min(f[i][j]3,f[i][k]2+a[k][j])=a*a*a=a3;//经过三条边的最短路径，经过二次floyd。f[i][j]3初值为∞

f[i][j]4=min(f[i][j]4,f[i][k]3+a[k][j])=a4;//经过四条边的最短路径，经过三次floyd,f[i][j]4初值为∞

    ...

f[i][j]k=min(f[i][j]k,f[i][k]k-1+a[k][j])=ak;//经过k条边的最短路径，经过K-1次floyd,f[i][j]k初值为∞ 而floyd的时间复杂度为O（n3）,则从的时间复杂度为O（Kn3），非常容易超时。 所以我们可以用快速幂来完成。

f[i][j]r+p=min(f[i][j]r+p,f[i][k]r+f[k][j]p)

程序：

//洛谷2886 //（1）对角线不能设置为0，否则容易自循环。（2）数组要放在主程序外面 //（3）K条边，不一定是最简路 #include
       
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