poj3613 Cow Relays

题目

题目传送门

Description

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 ≤ I_1i ≤ 1,000; 1 ≤ I_2i ≤ 1,000), each of which is the termination for at least two trails. The cows know the length_i of each trail (1 ≤ length_i  ≤ 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.

Input

* 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: length_i , I_1i , and I_2i

Output

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

Sample Input

2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9

Sample Output

10

翻译

题目描述

给定T条边的无向图，点的编号为1~1000之间的整数。求从起点S到终点E恰好经过N条边（可重复）的最短路。 
2<=T<=100 
2<=N<=1000000 

输入

第一行输入四个数，分别表示N，T，S，E 
以下T行分别表示每条路径w，a，b，w表示路径长度 

输出

输出从S到T经过所有牛的最短路

样例输入

2 6 6 4
11 4 6
4 4 8
8 4 9
6 6 8
2 6 9
3 8 9

样例输出

10

分析

由于N很大并且求的是“N最短路”所以：

这道题可以用floyd的一部分思路和快速幂的精髓。

但在写完之后你会发现编（zuo）号（zhe）是（te）乱（bie）的（hen），这时我们需要我们需要用到哈希。

代码

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int n,t,s,e,a[1001][1001],o[1001][1001],ans[1001][1001],vis[1001],p[1001],number[1001],minn[1001][1001],num;
inline int put(int x)//哈希 
{
    if(!vis[x]) p[++num]=x,vis[x]=1,number[x]=num;
	return number[x];
}
inline void floyd()
{
	memcpy(minn,ans,sizeof(minn));
	memset(ans,999999,sizeof(ans));
	for(int k=1;k<=num;k++)
	    for(int i=1;i<=num;i++)
	        for(int j=1;j<=num;j++)
	            ans[i][j]=min(minn[i][k]+a[k][j],ans[i][j]);
}
inline void change()
{
	memcpy(o,a,sizeof(o));
	memset(a,999999,sizeof(a));
	for(int k=1;k<=num;k++)
	    for(int i=1;i<=num;i++)
	        for(int j=1;j<=num;j++)
	            a[i][j]=min(o[i][k]+o[k][j],a[i][j]);
}
inline int sq(int b)//快速幂的精髓 
{
	while(b)
	{
		if(b&1) floyd();
		change();
		b>>=1;
	}
	return ans[number[s]][number[e]];
}
int main()
{
	memset(a,999999,sizeof(a));
	memset(ans,999999,sizeof(ans));//赋初值（这里本人建议用999999） 
	cin>>n>>t>>s>>e;
	for(int i=1;i<=t;i++)
	{
		int x,y,z;
		cin>>z>>x>>y;
		a[put(y)][put(x)]=a[put(x)][put(y)]=z;//用哈希节约时间、空间 
	} 
	for(int i=1;i<=num;i++) ans[i][i]=0;//自己到自己为0 
	cout<<sq(n);
	return 0;
}