题目大意：给定一些点之间的位置关系，求两个点之间的曼哈顿距离

此题土豪题，不过POJ也有一道同样的题，可以刷一下

别被题目坑到了，这题不强制在线，把询问离线处理即可

然后就是带权并查集的问题了。。。将权值设为方向向量，重载+和-，按照正常权值并查集做就行了

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 40400
using namespace std;
struct abcd{
	int x,y;
	abcd(){}
	abcd(int X,int Y):x(X),y(Y){}
	abcd operator + (const abcd &Y) const
	{
		return abcd( x+Y.x , y+Y.y );
	}
	abcd operator - (const abcd &Y) const
	{
		return abcd( x-Y.x , y-Y.y );
	}
}f[M];
struct operation{
	int x,y;
	abcd temp;
}operations[M];
struct query{
	int x,y,z,pos;
	bool operator < (const query &Y) const
	{
		return z < Y.z ;
	}
}queries[10100];
int n,m,q,fa[M],ans[10100];
int Distance(abcd x)
{
	return abs(x.x)+abs(x.y);
}
int Find(int x)
{
	if(!fa[x]||fa[x]==x)
		return fa[x]=x;
	int y=fa[x];
	fa[x]=Find(fa[x]);
	f[x]=f[y]+f[x];
	return fa[x];
}
int main()
{
	int i,j,x,y,z;
	char p[10];
	cin>>n>>m;
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%d%s",&operations[i].x,&operations[i].y,&z,p);
		switch(p[0])
		{
			case 'E':operations[i].temp=abcd(z,0);break;
			case 'W':operations[i].temp=abcd(-z,0);break;
			case 'N':operations[i].temp=abcd(0,z);break;
			case 'S':operations[i].temp=abcd(0,-z);break;
		}
	}
	cin>>q;
	for(i=1;i<=q;i++)
		scanf("%d%d%d",&queries[i].x,&queries[i].y,&queries[i].z),queries[i].pos=i;
	sort(queries+1,queries+q+1);
	for(i=1,j=1;i<=q;i++)
	{
		for(;j<=queries[i].z;j++)
		{
			int x=operations[j].x;
			int y=operations[j].y;
			int fx=Find(x),fy=Find(y);
			fa[fy]=fx;
			f[fy]=f[x]-f[y]+operations[j].temp;
		}
		int x=queries[i].x;
		int y=queries[i].y;
		if( Find(x)!=Find(y) )
			ans[queries[i].pos]=-1;
		else
			ans[queries[i].pos]=Distance(f[x]-f[y]);
	}
	for(i=1;i<=q;i++)
		printf("%d\n",ans[i]);	
}

BZOJ 3362 Navigation Nightmare 带权并查集