hdu_5044_Tree(树链剖分)

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=5044

题意：给一棵树，在点和边上操作

题解：树链剖分，剖完后用树状数组维护即可，因为只有加减操作，连树状的部分都不用写，最后要注意当n等于1的情况

 #include<cstdio>
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #define F(i,a,b) for(int i=a;i<=b;++i)

 const int N=;char op[];
 int t,ic=,n,m,x,y,c,egu[N],egv[N],g[N],nxt[*N],v[*N],ed,dep[N],sz[N],fa[N],hs[N],tid[N],top[N],idx,fid[N],fid2[N];
 __int64 a[N],b[N],ans1[N],ans2[N],all;

 inline void adg(int x,int y){v[++ed]=y,nxt[ed]=g[x],g[x]=ed;}

 void dfs1(int u,int pre){
     sz[u]=,fa[u]=pre,hs[u]=,dep[u]=dep[pre]+;
     for(int i=g[u];i;i=nxt[i]){
         int vv=v[i];
         if(vv!=pre){
             dfs1(vv,u);
             if(sz[vv]>hs[u])hs[u]=vv;
             sz[u]+=sz[vv];
         }
     }
 }

 void dfs2(int u,int tp){
     tid[u]=++idx,fid[idx]=u,top[u]=tp;
     if(hs[u])dfs2(hs[u],tp);
     for(int i=g[u];i;i=nxt[i]){
         int vv=v[i];
         if(vv!=fa[u]&&vv!=hs[u])dfs2(vv,vv);
     }
 }

 void up(__int64*a,int x,int y,int c,int k){
     int fx=top[x],fy=top[y];
     while(fx!=fy){
         if(dep[fx]>dep[fy])
             a[tid[fx]]+=c,a[tid[x]+]-=c,x=fa[fx],fx=top[x];
         else a[tid[fy]]+=c,a[tid[y]+]-=c,y=fa[fy],fy=top[y];
     }
     if(dep[x]>dep[y])x=x^y,y=x^y,x=x^y;
     a[tid[x]+k]+=c,a[tid[y]+]-=c;
 }

 int main(){
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&m);
         F(i,,n)g[i]=;ed=;
         F(i,,n-)scanf("%d%d",egu+i,egv+i),adg(egu[i],egv[i]),adg(egv[i],egu[i]);
         dfs1(,),idx=,dfs2(,);
         F(i,,n)a[i]=,b[i]=;all=;
         while(m--){
             scanf("%s%d%d%d",op,&x,&y,&c);
             if(op[]=='')up(a,x,y,c,);
             else up(b,x,y,c,);
         }
         F(i,,n-)if(dep[egu[i]]>=dep[egv[i]])fid2[tid[egu[i]]]=i;
         else fid2[tid[egv[i]]]=i;
         F(i,,n)all+=a[i],ans1[fid[i]]=all;all=;
         F(i,,n)all+=b[i],ans2[fid2[i]]=all;
         printf("Case #%d:\n",ic++);
         F(i,,n)printf("%I64d%c",ans1[i],i==n?'\n':' ');
         F(i,,n-)printf("%I64d%c",ans2[i],(i==n-?'\n':' '));
         if(n==)puts("");//PE
     }
     return ;
 }