[USACO2007NOVS] Cow Hurdles S

题目描述

Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hurdles.

Obviously, it is not very difficult for a cow to jump over several very short hurdles, but one tall hurdle can be very stressful. Thus, the cows are only concerned about the height of the tallest hurdle they have to jump over.

The cows' practice room has \(N (1 ≤ N ≤ 300) stations\), conveniently labeled \(1..N\). A set of \(M (1 ≤ M ≤ 25,000)\) one-way paths connects pairs of stations; the paths are also conveniently labeled \(1..M\). Path i travels from station Si to station Ei and contains exactly one hurdle of height \(H_i\) (\(1 ≤ H_i ≤ 1,000,000\)). Cows must jump hurdles in any path they traverse.

The cows have \(T\) (1 ≤ \(T\) ≤ 40,000) tasks to complete. Task \(i\) comprises two distinct numbers, Ai and Bi (\(1 ≤ A_i ≤ N\);$ 1 ≤ B_i ≤ N$), which connote that a cow has to travel from station \(A_i\) to station \(B_i\) (by traversing over one or more paths over some route). The cows want to take a path the minimizes the height of the tallest hurdle they jump over when traveling from \(A_i\) to \(B_i\) . Your job is to write a program that determines the path whose tallest hurdle is smallest and report that height.

输入格式

* Line 1: Three space-separated integers: N, M, and T

* Lines 2..M+1: Line i+1 contains three space-separated integers: \(S_i\) , \(E_i\) , and \(H_i\)

* Lines M+2..M+T+1: Line i+M+1 contains two space-separated integers that describe task i: \(A_i\) and \(B_i\)

输出格式

* Lines \(1..T\): Line \(i\) contains the result for task \(i\) and tells the smallest possible maximum height necessary to travel between the stations. Output -1 if it is impossible to travel between the two stations.

5 6 3
1 2 12
3 2 8
1 3 5
2 5 3
3 4 4
2 4 8
3 4
1 2
5 1

样例输出 #1

4
8
-1

方法类似最短路。首先可以用floyd预处理出两点间路径最大值的最小值（也就是以最大值作为更新Floyd的过程），然后回答询问即可。

#include<bits/stdc++.h>
using namespace std;
const int N=305;
int n,m,t,u,v,w;
int dp[N][N];
int main()
{
	memset(dp,0x7f,sizeof(dp));
	scanf("%d%d%d",&n,&m,&t);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		dp[u][v]=min(dp[u][v],w);
	}
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				dp[i][j]=min(dp[i][j],max(dp[i][k],dp[k][j]));
	while(t--)
	{
		scanf("%d%d",&u,&v);
		if(dp[u][v]>1000000)
			printf("-1\n");
		else
			printf("%d\n",dp[u][v]);
	}
	return 0;
}

优化1：可以将Floyd改为跑n遍Dijkstra，因为一个位置如果现在是最小的，那么别的点再走一条边取一个max最多也就会更这个位置一样，所以dijkstra的贪心仍然符合。
优化2：参考货车运输