BZOJ 3362 Navigation Nightmare

一道带权并查集题目.

带权并查集的重点是信息的合并.

这类题出现得并不多,练习一下.

 #include<bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define FILE "dealing"
 #define up(i,j,n) for(int i=j;i<=n;i++)
 #define db long double
 #define pii pair<int,int>
 #define pb push_back
 #define mem(a,L) memset(a,0,sizeof(int)*(L+1))
 template<class T> inline bool cmin(T& a,T b){return a>b?a=b,true:false;}
 template<class T> inline bool cmax(T& a,T b){return a<b?a=b,true:false;}
 template<class T> inline T squ(T a){return a*a;}
 const ll maxn=+,inf=1e9+,mod=,M=;
 ll read(){
     ll x=,f=,ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<='')x=(x<<)+(x<<)+ch-'',ch=getchar();
     return x*f;
 }
 int l[maxn],r[maxn],fa[maxn];
 int getfa(int x){
     if(fa[x]==x)return x;
     int y=getfa(fa[x]);
     if(y==fa[x])return y;
     l[x]+=l[fa[x]],r[x]+=r[fa[x]];
     fa[x]=y;
     return y;
 }
 int n,m,Q;
 int f1[maxn],f2[maxn],d[maxn],op[maxn];
 int xx[maxn],yy[maxn],t[maxn],ans[maxn];
 vector<int> linkk[maxn];
 int main(){
     freopen(FILE".in","r",stdin);
     freopen(FILE".out","w",stdout);
     n=read(),m=read();
     char ch;
     up(i,,m){
         f1[i]=read(),f2[i]=read(),d[i]=read();
         scanf(" %c",&ch);
         if(ch=='E')op[i]=;
         else if(ch=='W')op[i]=;
         else if(ch=='N')op[i]=;
         else if(ch=='S')op[i]=;
     }
     Q=read();
     up(i,,Q){
         xx[i]=read(),yy[i]=read(),t[i]=read();
         linkk[t[i]].push_back(i);
     }
     up(i,,n)fa[i]=i;
     up(i,,m){
         int x=f1[i],y=f2[i],fx=getfa(x),fy=getfa(y);
         if(fx!=fy){
             if(fx!=x)l[fx]=-l[x],r[fx]=-r[x],l[x]=r[x]=,fa[fx]=x,fa[x]=x;
             if(fy!=y)l[fy]=-l[y],r[fy]=-r[y],l[y]=r[y]=,fa[fy]=y,fa[y]=y;
             fa[x]=y;
             if(op[i]==)l[x]=d[i];
             else if(op[i]==)l[x]=-d[i];
             else if(op[i]==)r[x]=d[i];
             else r[x]=-d[i];
         }
         for(int j=;j<linkk[i].size();j++){
             int t=linkk[i][j];
             int x=xx[t],y=yy[t];
             int fx=getfa(x),fy=getfa(y);
             if(fx!=fy)ans[t]=-;
             else ans[t]=abs(l[x]-l[y])+abs(r[x]-r[y]);
         }
     }
     up(i,,Q)printf("%d\n",ans[i]);
     return ;
 }