hdu3592 World Exhibition--单源最短路径&差分约束-白红宇的个人博客

发布日期：2021-10-03 20:32:11 浏览次数：2 分类：技术文章

本文共 3202 字，大约阅读时间需要 10 分钟。

原题链接：

一：原题内容

Problem Description

Nowadays, many people want to go to Shanghai to visit the World Exhibition. So there are always a lot of people who are standing along a straight line waiting for entering. Assume that there are N (2 <= N <= 1,000) people numbered 1..N who are standing in the same order as they are numbered. It is possible that two or more person line up at exactly the same location in the condition that those visit it in a group.

There is something interesting. Some like each other and want to be within a certain distance of each other in line. Some really dislike each other and want to be separated by at least a certain distance. A list of X (1 <= X <= 10,000) constraints describes which person like each other and the maximum distance by which they may be separated; a subsequent list of Y constraints (1 <= Y <= 10,000) tells which person dislike each other and the minimum distance by which they must be separated.

Your job is to compute, if possible, the maximum possible distance between person 1 and person N that satisfies the distance constraints.

Input

First line: An integer T represents the case of test.

The next line: Three space-separated integers: N, X, and Y.

The next X lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= N. Person A and B must be at most C (1 <= C <= 1,000,000) apart.

The next Y lines: Each line contains three space-separated positive integers: A, B, and C, with 1 <= A < B <= C. Person A and B must be at least C (1 <= C <= 1,000,000) apart.

Output

For each line: A single integer. If no line-up is possible, output -1. If person 1 and N can be arbitrarily far apart, output -2. Otherwise output the greatest possible distance between person 1 and N.

Sample Input

14 2 11 3 82 4 152 3 4

Sample Output

二：分析理解

一次AC的题！

对于X行的来说：B-A<=C

对于Y行的来说：B-A>=C

最后：(i) -( i-1)>=0，其中2<=i<=N

因为题意要求是最大值所以转化一下就是：

B-A<=C;

A-B<=-C;

(i-1)-(i)<=0;

三：AC代码

#include
   
        #include
    
       #include
     
        using namespace std;struct Edge{	int v;	int w;	int next;};Edge edge[30005];int cnt[1005];int sta[1005];bool visited[1005];int dis[1005];int head[1005];int num;int T;int N, X, Y;int A, B, C;void AddEdge(int u, int v, int w){	edge[num].v = v;	edge[num].w = w;	edge[num].next = head[u];	head[u] = num++;}int Spfa(){	dis[1] = 0;	int top = 1;	sta[top] = 1;	visited[1] = 1;	while (top)	{		int u = sta[top--];		visited[u] = 0;		for (int i = head[u]; i != -1; i = edge[i].next)		{			int v = edge[i].v;			if (dis[v] > dis[u] + edge[i].w)			{				dis[v] = dis[u] + edge[i].w;				if (!visited[v])				{					sta[++top] = v;					visited[v] = 1;					cnt[v]++;				}				if (cnt[v] > N)					return -1;			}		}	}	if (dis[N] == 99999999)		return -2;	return dis[N];}int main(){	scanf("%d", &T);	while (T--)	{		num = 0;		memset(head, -1, sizeof(head));		memset(cnt, 0, sizeof(cnt));		memset(visited, 0, sizeof(visited));		fill(dis, dis + 1005, 99999999);		scanf("%d%d%d", &N, &X, &Y);		while (X--)		{			scanf("%d%d%d", &A, &B, &C);			AddEdge(A, B, C);		}		while (Y--)		{			scanf("%d%d%d", &A, &B, &C);			AddEdge(B, A, -C);		}		for (int i = 2; i <= N; i++)			AddEdge(i, i - 1, 0);		printf("%d\n", Spfa());	}	return 0;}

转载地址：https://blog.csdn.net/LaoJiu_/article/details/51043451 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！

上一篇：hdu3440 House Man--单源最短路径&差分约束

下一篇：hdu1529 Cashier Employment--单源最短路径&差分约束

发表评论

关于作者

喝酒易醉，品茶养心，人生如梦，品茶悟道，何以解忧？唯有杜康！

-- 愿君每日到此一游！

发表评论

最新留言

关于作者

推荐文章