POJ3613 Cow Relays 矩阵快速幂 Floyd

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Cow Relays

Time Limit: 1000MS $ \quad $ Memory Limit: 65536K 　 ### Description

For their physical fitness program,
$ N (2 \le N \le 1,000,000) $ cows have decided to run a relay race
using the $ T (2 \le T \le 100) $ cow trails throughout the pasture.
　
Each trail connects two different intersections $ (1 \le I_{1i} \le 1,000; 1 \le I_{2i} \le 1,000) $ ,
each of which is the termination for at least two trails.
The cows know the $ length_i $ of each trail $ (1 \le length_i \le 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

　

Source

USACO 2007 November Gold

　

题目大意

• 给定一张由 $ T (2 \le T \le 100 ) $ 条边构成的无向图，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 (2 \le N \le 1000000) $ 条边（可重复经过）的最短路。

　

题解

• 点的标号离散化到 $ 1 ~ P $

• 若矩阵 $ A^m $ 保存任意两点之间恰好经过 $ m $ 条边的最短路，则：

\[\forall i,j \in [1,P], \quad (A^{r+m})[i,j]=min_{1 \le k \le P}((A^r)[i,k]+(A^m)[k,j]) \]

• 上式其实等价于一个关于 $ min $ 与加法运算的广义“矩阵乘法”。

• $ (A^N)[S,E] $ 即为本题的答案。

　

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,t,s,e,mp[1010],cnt;
struct data{
	int a[205][205];
	data operator *(data &x){
		data c; memset(c.a,0x3f,sizeof(c.a));
		for(int i=1;i<=cnt;++i)
			for(int j=1;j<=cnt;++j)
				for(int k=1;k<=cnt;++k)
					c.a[i][j]=min(c.a[i][j],a[i][k]+x.a[k][j]);
		return c;
	}
}st,ans;
void qpow(){
	ans=st; --n;
	while(n){
		if(n&1) ans=ans*st;
		st=st*st;
		n>>=1;
	}
}
int main(){
	scanf("%d %d %d %d",&n,&t,&s,&e);
	memset(st.a,0x3f,sizeof(st.a));
	for(int w,u,v,i=1;i<=t;++i){
		scanf("%d %d %d",&w,&u,&v);
		if(mp[u]) u=mp[u]; else u=mp[u]=++cnt;
		if(mp[v]) v=mp[v]; else v=mp[v]=++cnt;
		st.a[u][v]=st.a[v][u]=w;
	}
	qpow();
	printf("%d",ans.a[mp[s]][mp[e]]);
	return 0;
}
/*
Problem: 3613

User: potremz
Memory: 788K

Time: 297MS
Language: C++

Result: Accepted
*/