差分约束——HDOJ 3592

HDOJ 3592 World Exhibition

/*
HDOJ 3592
典型的差分约束问题
x <= a-b <= y
a-b >= x
b-a <= -x
全部转化为<=的形式，求上界，最短路
*/

#include <iostream>
#include <vector>
#include <deque>
using namespace std;

#define INF 1000009
#define MAXN 1005

struct edge
{
	int to;
	int weight;
};

vector <edge> adjmap[MAXN];
bool in_queue[MAXN];  //顶点是否在队列中
int in_sum[MAXN];  //顶点入队次数
int dist[MAXN];  //源点到各点的最短路径

int n;

bool SPFA(int source)
{
	deque <int> dq;
	int i,j,x,to;
	for(i=1;i<=n;i++)
	{
		in_sum[i]=0;
		in_queue[i]=false;
		dist[i]=INF;
	}
	dq.push_back(source);
	in_sum[source]++;
	in_queue[source]=true;
	dist[source]=0;
	
	while(!dq.empty())
	{
		x=dq.front();
		dq.pop_front();
		in_queue[x]=false;
		for(i=0;i<adjmap[x].size();i++)
		{
			to=adjmap[x][i].to;
			if((dist[x] < INF)&&(dist[to] > dist[x]+adjmap[x][i].weight))
			{
				dist[to]=dist[x]+adjmap[x][i].weight;
				if(!in_queue[to])
				{
					in_queue[to]=true;
					in_sum[to]++;
					if(in_sum[to] == n)
						return false;
					if(!dq.empty())
					{
						if(dist[to] > dist[dq.front()])
							dq.push_back(to);
						else
							dq.push_front(to);
					}
					else
						dq.push_back(to);
				}
			}
		}
	}
	return true;
}

int main()
{
	int nCase,xn,yn,i,j,a,b,c;
	edge temp;
	cin>>nCase;
	while(nCase--)
	{
		cin>>n>>xn>>yn;
		for(i=0;i<=n;i++)
			adjmap[i].clear();
		for(i=0;i<xn;i++)
		{
			cin>>a>>b>>c;
			temp.to=b;
			temp.weight=c;
			adjmap[a].push_back(temp);
		}
		for(i=0;i<yn;i++)
		{
			cin>>a>>b>>c;
			temp.to=a;
			temp.weight= -c;
			adjmap[b].push_back(temp);
		}
		if(SPFA(1))
		{
			if(dist[n]!=INF)
				cout<<dist[n]<<endl;
			else
				cout<<-2<<endl;
		}
		else
			cout<<-1<<endl;
	}
	return 0;
}