POJ3169 Layout【差分约束】

Layout

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15418		Accepted: 7406

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.

Source

USACO 2005 December Gold

问题链接：POJ3169 Layout

问题描述：有N只奶牛（从1开始编号）。这N只奶牛按编号顺序站在一条直线上，奶牛们想和自己喜欢的奶牛离得近些，与自己不喜欢的奶牛离得远些，输入 N, ML,  MD，分别表示奶牛的总数，ML条限制：A B D表示奶牛A和B最多相距D(喜欢)。MD条限制：A B D表示奶牛A和B至少相距D（不喜欢）。注意多个奶牛可以站在同一个位置。问在满足限制的条件下1和N的距离的最大值。

解题思路：差分约束求最大值，用最短路。设d[i]表示奶牛i与奶牛1的距离，注意有隐含限制条件：由于奶牛们是按编号顺序排列的，且多个奶牛可以站在同一个位置，所以有d[i]-d[i-1]>=0。

AC的C++程序：

#include<iostream>
#include<cstring>
#include<vector>
#include<queue>

using namespace std;

const int N=1010;
const int INF=0x3f3f3f3f;

struct Edge{
	int v,w;
	Edge(int v,int w):v(v),w(w){}
};

vector<Edge>g[N];
int dist[N];
bool vis[N];
int cnt[N];
int n;//结点数 

bool spfa(int s)
{
	memset(vis,false,sizeof(vis));
	memset(cnt,0,sizeof(cnt));
	memset(dist,INF,sizeof(dist));
	dist[s]=0;
	cnt[1]=1;
	vis[s]=true;
	queue<int>q;
	q.push(s);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=false;
		for(int i=0;i<g[u].size();i++)
		{
			int v=g[u][i].v;
			int w=g[u][i].w;
			if(dist[v]>dist[u]+w)
			{
				dist[v]=dist[u]+w;
				if(!vis[v])
				{
					vis[v]=true;
					q.push(v);
					cnt[v]++;
					if(cnt[v]>n)
					  return false;//存在环 
				}
			}
		}
	}
	return true; 
}

int main()
{
	int l,d,a,b,w;
	while(~scanf("%d%d%d",&n,&l,&d))
	{
		for(int i=0;i<N;i++)
		  g[i].clear();
		while(l--)
		{
			scanf("%d%d%d",&a,&b,&w);
			g[a].push_back(Edge(b,w));
		}
		while(d--)
		{
			scanf("%d%d%d",&a,&b,&w);
			g[b].push_back(Edge(a,-w));
		}
		//隐含限制条件
		for(int i=2;i<=n;i++)
		  g[i].push_back(Edge(i-1,0));
		bool flag=spfa(1);
		if(flag)
		{
			if(dist[n]==INF)//1到n不可达 
			  printf("-2\n");
			else
			  printf("%d\n",dist[n]);
		}
		else//存在负环 
		  printf("-1\n"); 
	}
	return 0;
}