1890. Money out of Thin Air（线段树 dfs转换区间）

1890

将树的每个节点都转换为区间的形式 然后再利用线段树对结点更新 这题用了延迟标记 相对普通线段树 多了dfs的转换 把所要求的转换为某段区间

RE了N次 最后没办法了 记得有个加栈的语句 拿来加上A了。。

 #pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<stdlib.h>
 #include<vector>
 #include<algorithm>
 using namespace std;
 #define N 50010
 #define LL long long
 vector<int>w[N];
 LL lleft[N],rright[N],po[N],c[N],cnt;
 LL o[N];
 LL s[N<<],te[N<<];
 void up(int w)
 {
     s[w] = s[w<<]+s[w<<|];
 }
 void build(int l,int r,int w)
 {
     if(l==r)
     {
         s[w] = c[l];
         return ;
     }
     int m = (l+r)>>;
     build(l,m,w<<);
     build(m+,r,w<<|);
     up(w);
 }
 void pushdown(int w,int d)
 {
     if(te[w])
     {
         te[w<<] += te[w];
         te[w<<|]+=te[w];
         s[w<<] += (d-d/)*te[w];
         s[w<<|]+=d/*te[w];
         te[w] = ;
     }
 }
 void update(int a,int b,LL v,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         te[w] += v;
         s[w] += v*(r-l+);
         return ;
     }
     pushdown(w,r-l+);
     int m = (l+r)>>;
     if(a<=m)
     update(a,b,v,l,m,w<<);
     if(b>m)
     update(a,b,v,m+,r,w<<|);
     up(w);
 }
 LL query(int a,int b,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         return s[w];
     }
     pushdown(w,r-l+);
     int m = (l+r)>>;
     LL re =;
     if(a<=m)
     re+=query(a,b,l,m,w<<);
     if(b>m)
     re+=query(a,b,m+,r,w<<|);
     return re;
 }
 void dfs(int u)
 {
     int i;
     for(i =  ; i < (int)w[u].size() ; i++)
     {
         int v = w[u][i];
         dfs(v);
         if(lleft[u]==-)
         lleft[u] = lleft[v];
         else
         lleft[u] = min(lleft[v],lleft[u]);
         if(rright[u]==-)
         rright[u] = rright[v];
         else
         rright[u] = max(rright[u],rright[v]);
     }
     cnt++;
     c[cnt] = o[u];
     po[u] = cnt;
     if(lleft[u]==-)
     lleft[u] = cnt;
     else
     lleft[u] = min(lleft[u],cnt);
     if(rright[u]==-)
     rright[u] = cnt;
     else
     rright[u] = max(rright[u],cnt);
 }
 int main()
 {
     int n,q,s0,x,u,v,i;
     char ss[];
     memset(lleft,-,sizeof(lleft));
     memset(rright,-,sizeof(rright));
     scanf("%d%d%d",&n,&q,&s0);
     o[] = s0;
     for(i =  ; i <= n ; i++)
     w[i].clear();
     for(i = ; i <= n ; i++)
     {
         scanf("%d%d",&u,&v);
         u++;
         w[u].push_back(i);
         o[i] = v;
     }
     dfs();
     build(,n,);
     while(q--)
     {
         scanf("%s%d%d%d",ss,&x,&u,&v);
         x++;
         if(ss[]=='e')
         {
             int pp = po[x];
             LL tx = query(pp,pp,,n,);
             if(tx<u)
             update(pp,pp,v,,n,);
         }
         else if(ss[]=='d')
         {
             int ll = lleft[x];
             int rr = rright[x];
             int nu = rr-ll+;
             LL tx = query(ll,rr,,n,);
             if(1.0*tx/nu<u)
             update(ll,rr,v,,n,);
         }
     }
     for(i = ; i <= n ; i++)
     printf("%lld\n",query(po[i],po[i],,n,));
     return ;
 }