BZOJ 3362: [Usaco2004 Feb]Navigation Nightmare 导航噩梦

Description

给你每个点与相邻点的距离和方向,求两点间的曼哈顿距离. (n leqslant 4 imes 10^4) .

Sol

加权并查集.

像向量合成一样合并就可以了,找 (f[x]) 的时候需要先记录现在的父节点,然后更新他新的父节点.

Code

/**************************************************************
    Problem: 3362
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:80 ms
    Memory:3712 kb
****************************************************************/
 
#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
 
typedef pair< int,int > pr;
#define abs(x) ((x)<0 ? -(x) : (x) )
#define mpr make_pair
#define debug(a) cout<<#a<<"="<<a<<" "
const int N = 40005;
 
int n,m,k;
int f[N],ans[N];pr g[N];
struct Q{ int a,b,c,d; }q[N],qq[N];
 
inline int in(int x=0,char ch=getchar()){ while(ch>'9' || ch<'0') ch=getchar();
    while(ch>='0' && ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x ;}
inline char read(char ch=getchar()){ while(ch>'Z' || ch<'A') ch=getchar();return ch; }
pr operator + (const pr &a,const pr &b){ return mpr(a.first+b.first,a.second+b.second); }
pr operator += (pr &a,const pr &b){ return a=a+b; }
pr operator - (const pr &a,const pr &b){ return mpr(a.first-b.first,a.second-b.second); }
pr operator -= (pr &a,const pr &b){ return a=a-b; } 
int cmp1(const Q &x,const Q &y){ return x.c<y.c; }
int cmp2(const Q &x,const Q &y){ return x.d<y.d; }
int find(int x){
    if(f[x] == x) return x;
    int t=f[x];f[x]=find(f[x]);
    g[x]+=g[t];return f[x];
}
void work(int u,int v,int w,char d){
    int r1=find(u),r2=find(v);
//  debug(r1),debug(r2)<<endl;
//  cout<<u<<" "<<v<<" "<<w<<" "<<d<<endl;
//  cout<<"qwq"<<endl;
    if(r1!=r2) switch(d){
        case 'N':f[r1]=r2,g[r1]=g[v]+mpr(0,w)-g[u];break;
        case 'W':f[r1]=r2,g[r1]=g[v]+mpr(-w,0)-g[u];break;
        case 'S':f[r1]=r2,g[r1]=g[v]+mpr(0,-w)-g[u];break;
        default:f[r1]=r2,g[r1]=g[v]+mpr(w,0)-g[u];break;
    }
//  for(int i=1;i<=n;i++) cout<<find(i)<<":("<<g[i].first<<","<<g[i].second<<") ";cout<<endl;
}
int main(){
//  freopen("in.in","r",stdin);
    n=in(),m=in();
    for(int i=1;i<=n;i++) f[i]=i;
    for(int i=1;i<=m;i++) qq[i].a=in(),qq[i].b=in(),qq[i].c=in(),qq[i].d=read();
    k=in();for(int i=1;i<=k;i++) q[i].a=in(),q[i].b=in(),q[i].c=in(),q[i].d=i;
    sort(q+1,q+k+1,cmp1);
//  for(int i=1;i<=m;i++) work(qq[i].a,qq[i].b,qq[i].c,qq[i].d);
//  for(int i=1;i<=n;i++) cout<<find(i)<<" ("<<g[i].first<<","<<g[i].second<<")
";
     
    for(int i=1,j=1,u,v,r1,r2;i<=k;i++){
        for(;j<=q[i].c && j<=n;++j) work(qq[j].a,qq[j].b,qq[j].c,qq[j].d);
        u=q[i].a,v=q[i].b,r1=find(u),r2=find(v);
        if(r1==r2){
//          debug(u),debug(v),debug(g[u].first),debug(g[v].first),debug(g[u].second),debug(g[v].second)<<endl;
            ans[q[i].d]=abs(g[u].first-g[v].first)+abs(g[u].second-g[v].second);
        }else ans[q[i].d]=-1;
    }
//  for(int i=1;i<=n;i++) cout<<find(i)<<" ("<<g[i].first<<","<<g[i].second<<")
";
    for(int i=1;i<=k;i++) printf("%d
",ans[i]);
    return 0;
}