zoj 2866 Overstaffed Company

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1866

题目大意:一颗多叉树中,每个结点有一个权值,求每个节点有多少个其子孙节点的权值大于其权值.

题目思路:用树状数组快速求,求某一节点时,之前必然要把他的子节点都插到树状数组,所以要用遍历完子树才能求值,但是遍历子树之前也需先求值,因为之前有其他的节点插到树状数组了.

代码:

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <time.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;//#define ull unsigned __int64
//#define ll __int64
//#define ull unsigned long long
//#define ll long long
#define son1 New(p.xl,xm,p.yl,ym),(rt<<2)-2
#define son2 New(p.xl,xm,min(ym+1,p.yr),p.yr),(rt<<2)-1
#define son3 New(min(xm+1,p.xr),p.xr,p.yl,ym),rt<<2
#define son4 New(min(xm+1,p.xr),p.xr,min(ym+1,p.yr),p.yr),rt<<2|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define middle (l+r)>>1
#define MOD 1000000007
#define esp (1e-8)
const int INF=0x3F3F3F3F;
const double DINF=10000.00;
//const double pi=acos(-1.0);
const int N=50010;
int n,m;
int sum[N],val[N],hash[N],ret[N];
vector<int>son[N];int bs(int key,int size,int A[]){int l=0,r=size-1,mid;while(l<=r){mid=middle;if(key>A[mid]) l=mid+1;else if(key<A[mid]) r=mid-1;else return mid;}return -1;
}int lowbit(int x){return x&(-x);}void Add(int x,int c){while(x<=m) sum[x]+=c,x+=lowbit(x);
}int Sum(int x){int r=0;while(x>0) r+=sum[x],x-=lowbit(x);return r;
}void Query(int rt){int i,pos=bs(val[rt],m,hash)+1;int tmp=Sum(m)-Sum(pos),len=son[rt].size();for(i=0;i<len;i++) Query(son[rt][i]);ret[rt]=Sum(m)-Sum(pos)-tmp;Add(pos,1);
}int main(){//freopen("1.in","r",stdin);//freopen("1.out","w",stdout);int i,j,k;//int T,cas;scanf("%d",&T);for(cas=1;cas<=T;cas++)while(~scanf("%d",&n)){for(i=0;i<n;i++) son[i].clear();for(i=1;i<n;i++){scanf("%d",&k);son[k].push_back(i);}for(i=0;i<n;i++){scanf("%d",&val[i]);hash[i]=val[i];}sort(hash,hash+n);for(i=m=1;i<n;i++) if(hash[i]!=hash[i-1]) hash[m++]=hash[i];memset(sum,0,sizeof(sum));Query(0);for(i=0;i<n-1;i++) printf("%d ",ret[i]);printf("%d\n",ret[i]);}return 0;
}

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