POJ3268_Silver Cow Party_队列优化的Dijkstra

最新推荐文章于 2022-03-13 20:30:09 发布

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最新推荐文章于 2022-03-13 20:30:09 发布

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分类专栏：最短路文章标签： poj 题解

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Silver Cow Party

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21567		Accepted: 9843

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i requires T_i (1 ≤ T_i ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively: N, M, and X
Lines 2.. M+1: Line i+1 describes road i with three space-separated integers: A_i, B_i, and T_i. The described road runs from farm A_i to farm B_i, requiring T_i time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

Sample Output

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

给定一个点X，求任意一点前往X的最短路和从X返回该点的最短路的和的最大值。

用dijkstra求两次最短路求和的最大值就好了

这个题是第一次用队列优化的dijkstra

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define f(x) x.first
#define s(x) x.second

using namespace std;
typedef pair<int, int> P;

const int maxn = 1010;
int Map [maxn][maxn];
int disf [maxn];
int dist [maxn];
int N, M, X;

void dijkstraf ()
{
	memset(disf, 0x3f, sizeof(disf) );
	priority_queue<P, vector<P>, greater<P> > qu;

	disf[X] = 0;
	qu.push(make_pair(0, X) );

	while(qu.size() ){
		int t = s(qu.top() ), v = f(qu.top() );
		qu.pop();
		if(v > disf[t] ) continue;

		for(int i= 1; i<= N; i++)
			if(disf[i] > disf[t] + Map[t][i]){
				disf[i] = disf[t] + Map[t][i];
				qu.push(make_pair(disf[i], i) );
			}
	}
}

void dijkstrat ()
{
	memset(dist, 0x3f, sizeof(dist) );
	priority_queue<P, vector<P>, greater<P> > qu;

	dist[X] = 0;
	qu.push(make_pair(0, X) );

	while(qu.size() ){
		int t = s(qu.top() ), v = f(qu.top() );
		qu.pop();
		if(v > dist[t] ) continue;

		for(int i= 1; i<= N; i++)
			if(dist[i] > dist[t] + Map[i][t]){
				dist[i] = dist[t] + Map[i][t];
				qu.push(make_pair(dist[i], i) );
			}
	}
}

int main ()
{
	scanf("%d %d %d", &N, &M, &X);

	memset(Map, 0x3f, sizeof(Map) );
	for(int i= 1; i<= N; i++)
		Map[i][i] = 0;

	for(int i= 1; i<= M; i++){
		int s, e, t;
		scanf("%d %d %d", &s, &e, &t);
		if(Map[s][e] > t) Map[s][e] = t;
	}

	dijkstraf();
	dijkstrat();

	int Max = 0;
	for(int i= 1; i<= N; i++){
		int t = disf[i] + dist[i];
		if(Max < t) Max = t;
	}

	printf("%d\n", Max);
}