hdu3592 差分约束问题

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.

经典差分约束问题

可知第一组数据中满足：B-A<=C

第二组满足B-A>=C -> A-B<=-C

又因为 [i]-[i-1]>=0

建图spfa即可得出结果

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
using namespace std;
const int N=1050;
const int inf=0x3f3f3f3f;
struct node
{
	int to,next,w;
}e[N*N/2];
int mp[N][N];
int k,vis[N],n;
int dis[N];
int num[N];
int head[N];
void add(int from,int to,int w)
{
	e[k].to=to;
	e[k].w=w;
	e[k].next=head[from];
	head[from]=k++;
}
int prim()
{
	memset(vis,0,sizeof(vis));
	memset(num,0,sizeof(num));
	dis[1]=0;
	for(int i=2;i<=n;i++)
		dis[i]=inf;
	queue<int> q;
	q.push(1);
	vis[1]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		++num[u];
		if(num[u]>n) return -1;
		for(int j=head[u];j!=-1;j=e[j].next)
		{
			int v=e[j].to,w=e[j].w;
			if(dis[v]>dis[u]+w)
			{
				dis[v]=dis[u]+w;
				if(!vis[v])
				{
					q.push(v);
					vis[v]=1;
				}
			}
		}
	}
	if(dis[n]==inf) return -2;
	return dis[n];
}
int main()
{
	int T,x,y,u,v,w;
	scanf("%d",&T);
	while(T--)
	{
		k=0;
		memset(head,-1,sizeof(head));
		scanf("%d %d %d",&n,&x,&y);
		while(x--)
		{
			scanf("%d %d %d",&u,&v,&w);
			add(u,v,w);
		}
		while(y--)
		{
			scanf("%d %d %d",&u,&v,&w);
			add(v,u,-w);
		}
		for(int i=2;i<=n;i++)
			add(i,i-1,0);
		printf("%d\n",prim());
	}
	return 0;
}