hdoj 3952 World Exhibition

World Exhibition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1383    Accepted Submission(s):
679

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

1

4 2 1

1 3 8

2 4 15

2 3 4

 

Sample Output

19

 

题意跟 poj3169  一模一样：分析点我！！！！

#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()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&n,&m,&s);
		init();
		getmap();
		spfa(1);
	}
	return 0;
}