Candies POJ-3159--最短路

最新推荐文章于 2021-10-19 20:46:54 发布

原创最新推荐文章于 2021-10-19 20:46:54 发布 · 289 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#最短路

最短路专栏收录该内容

11 篇文章

订阅专栏

本文探讨了一个关于幼儿园糖果公平分配的问题，通过构建最短路径模型，使用堆优化的Dijkstra算法和前向星建图技术，解决了如何在满足孩子间糖果数量差异约束下，实现糖果最大差距分配的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit: 1500MS

Memory Limit: 131072K

Description

During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and often compared the numbers of candies they got with others. A kid A could had the idea that though it might be the case that another kid B was better than him in some aspect and therefore had a reason for deserving more candies than he did, he should never get a certain number of candies fewer than B did no matter how many candies he actually got, otherwise he would feel dissatisfied and go to the head-teacher to complain about flymouse’s biased distribution.

snoopy shared class with flymouse at that time. flymouse always compared the number of his candies with that of snoopy’s. He wanted to make the difference between the numbers as large as possible while keeping every kid satisfied. Now he had just got another bag of candies from the head-teacher, what was the largest difference he could make out of it?

Input

The input contains a single test cases. The test cases starts with a line with two integers N and M not exceeding 30 000 and 150 000 respectively. N is the number of kids in the class and the kids were numbered 1 through N. snoopy and flymouse were always numbered 1 and N. Then follow M lines each holding three integers A, B and c in order, meaning that kid A believed that kid B should never get over c candies more than he did.

Output

Output one line with only the largest difference desired. The difference is guaranteed to be finite.

Sample Input

2 2
1 2 5
2 1 4

Sample Output

Hint

32-bit signed integer type is capable of doing all arithmetic.

题目大意：有N个人，让你求第一个人和第n个人的得到的糖果的最大差距。输入为a,b,c表示a认为b最多比他多c个糖果。

。。。看了一会儿题目，这不就是一个裸的最短路吗。。。假设a到b有很多路，我们必须选择最短的那一条，因为这样才能满足所有的猜测条件，这相当于取一个交集吧。。。于是这一题就没了，写个堆优化的Dij+前向星建图。。。一发就A了，没什么坑点。。。

以下是AC代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int inf = 1e8 + 10;
int n, m;
const int mac = 1e5 + 10;
struct node
{
	int to, w, next;
};
struct ns
{
	int id, d;
	bool operator<(const ns& a)const {
		return d > a.d;
	}
};
node eg[mac << 1];
int head[mac], dis[mac], num = 0, vis[mac];
void add(int u, int v, int w);
void Dij();
int main()
{
	scanf("%d%d", &n, &m);
	int u, v, w;
	memset(head, -1, sizeof(head));
	for (int i = 1; i <= m; i++) {
		scanf("%d%d%d", &u, &v, &w);
		add(u, v, w );
	}
	Dij();
	return 0;
}
void add(int u, int v, int w)
{
	eg[++num].to = v; eg[num].w = w; eg[num].next = head[u];
	head[u] = num;
}
void Dij()
{
	for (int i = 1; i <= n; i++) dis[i] = inf;
	memset(vis, 0, sizeof(vis));
	dis[1] = 0;
	priority_queue<ns>q;
	ns p;
	p.id = 1, p.d = 0;
	q.push(p);
	while (!q.empty()) {
		ns now = q.top();
		q.pop();
		int u = now.id;
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = head[u]; i != -1; i = eg[i].next) {
			int v = eg[i].to, w = eg[i].w;
			if (dis[v] > now.d + w) {
				dis[v] = now.d + w;
				ns p;
				p.d = now.d + w; p.id = v;
				q.push(p);
			}
		}
	}
	printf("%d\n", dis[n]);
}