poj 3169 Layout

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8610		Accepted: 4147

Description

Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and since they can be rather pushy, it is possible that two or more cows can line up at exactly the same location (that is, if we think of each cow as being located at some coordinate on a number line, then it is possible for two or more cows to share the same coordinate).

Some cows 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 ML (1 <= ML <= 10,000) constraints describes which cows like each other and the maximum distance by which they may be separated; a subsequent list of MD constraints (1 <= MD <= 10,000) tells which cows 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 cow 1 and cow N that satisfies the distance constraints.

Input

Line 1: Three space-separated integers: N, ML, and MD.

Lines 2..ML+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at most D (1 <= D <= 1,000,000) apart.

Lines ML+2..ML+MD+1: Each line contains three space-separated positive integers: A, B, and D, with 1 <= A < B <= N. Cows A and B must be at least D (1 <= D <= 1,000,000) apart.

Output

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

Sample Input

4 2 1
1 3 10
2 4 20
2 3 3

Sample Output

27

Hint

Explanation of the sample:

There are 4 cows. Cows #1 and #3 must be no more than 10 units apart, cows #2 and #4 must be no more than 20 units apart, and cows #2 and #3 dislike each other and must be no fewer than 3 units apart.

The best layout, in terms of coordinates on a number line, is to put cow #1 at 0, cow #2 at 7, cow #3 at 10, and cow #4 at 27.

 

题意：有n头牛，有些牛互相喜欢，他们想要离得越近越好，有的牛相互讨厌，他们想要离得越远越好，现在给你n头牛之间的关系，让你求出第一头牛与第n头之间的最小距离,当最小距离不存在，即存在负环时输出-1，当最短路为任意（即没有最短路，1到n之间不连通）时输出-2

输入：第一行输入三个数n，ML，MD，代表有n头牛，其中相互喜欢的有ML对，相互讨厌的有MD对，接下来ML行三个数a，b，c，代表a牛和b牛之间最多相隔 c 距离，接下来MD

行三个数a，b，c，代表a牛和b牛之间最少相隔 c 距离；

题解：根据题意找到约束条件，然后建图，由题意约束条件为：

一、相互喜欢的:b-a<=c    对应建边为add(a,b,c);

二、相互讨厌的：b-a>=c   ==>   a-b<=-c;    对应建边为add(b,a,-c);

#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 100000
#define INF 0x3f3f3f
using namespace std;
int head[MAX];
int n,m,s,ans;
int dis[MAX],vis[MAX];
int used[MAX];
struct node
{
	int u,v,w;
	int next;
}edge[MAX];
void add(int u,int v,int w)
{
	edge[ans].u=u;
	edge[ans].v=v;
	edge[ans].w=w;
	edge[ans].next=head[u];
	head[u]=ans++;
}
void init()
{
	ans=0;
	memset(head,-1,sizeof(head));
}
void getmap()
{
	int i,j;
	while(m--)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		add(a,b,c);
	}
	while(s--)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		add(b,a,-c);
	}
}
void spfa(int sx)
{
	int i,j;
	queue<int>q;
	memset(vis,0,sizeof(vis));
	memset(used,0,sizeof(used));
	for(i=1;i<=n;i++)
	    dis[i]=i==sx?0:INF;
	vis[sx]=1;
	used[sx]++;
	q.push(sx);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(i=head[u];i!=-1;i=edge[i].next)
		{
			int top=edge[i].v;
			if(dis[top]>dis[u]+edge[i].w)
			{
				dis[top]=dis[u]+edge[i].w;
				if(!vis[top])
				{
					vis[top]=1;
					q.push(top);
					used[top]++;
					if(used[top]>n)
					{
						printf("-1\n");
						return ;
					}
				}
			}
		}
	}
	if(dis[n]==INF)
	printf("-2\n");
	else
	printf("%d\n",dis[n]);
}
int main()
{
	while(scanf("%d%d%d",&n,&m,&s)!=EOF)
	{
		init();
		getmap();
		spfa(1);
	}
	return 0;
}