BZOJ 3362 Navigation Nightmare 带权并查集

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。只是POJ也有一道相同的题，能够刷一下

别被题目坑到了，这题不强制在线。把询问离线处理就可以

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-，依照正常权值并查集做即可了

#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;elseans[queries[i].pos]=Distance(f[x]-f[y]);}for(i=1;i<=q;i++)printf("%d\n",ans[i]);	
}