BOJ 395 Tree

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时间限制 7000 ms 内存限制 65536 KB

题目描述

Given a rooted tree with values assigned on each node, you're required to answer how many nodes on the path from the root to u, that have a value strictly greater than value[u]. Node 1 is the root.

输入格式

An integer T(T≤20) will exist in the first line, indicating the number of test cases. For each test case, the first line will be the number of tree nodes N(1≤N≤100000). The following (N−1) lines, each with a pair of integers (u,v), describe the edges of the tree. The values of nodes value[i](1≤value[i]≤109,1≤i≤N) will be put down on the next line.

输出格式

Output the answer for each node from node 1 to node N, one per line.

输入样例

141 21 34 24 1 5 0

输出样例

010 2

解题思路：

先将每一个节点离散化，再用一个数组C[x]，记录value值为x的节点的个数（在这里我使用的是树状数组，貌似用一般的数组好像会超时），然后在dfs的同时，记录该点value值的C数组以及该点前面的大于它的value值的点的个数，回溯的时候再回复C数组。

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define lowbit(x) x&(-x)using namespace std;const int MAXN=100010;vector<int> head[MAXN];int n,ans[MAXN];bool vis[MAXN];int c[MAXN];struct Con{int va,num;}a[MAXN];int tot=0,now[MAXN];//now表述压缩过的value，角标相同void update(int i,int v){for(;i<=tot;i+=lowbit(i))c[i]+=v;}int getsum(int i){int sum=0;for(;i>0;i-=lowbit(i))sum+=c[i];return sum;}void add(int a,int b){    head[a].push_back(b);    head[b].push_back(a);}void dfs(int x){vis[x]=true;update(now[x],1);    ans[x]=getsum(tot)-getsum(now[x]);for(int i=0;i<head[x].size();i++)    {        int temp=head[x][i];        if(!vis[temp])            dfs(temp);    }    update(now[x],-1);    return ;}bool cmp(Con a,Con b){    return a.va<b.va;}void compress(){    sort(a+1,a+n+1,cmp);    for(int i=1;i<=n;i++)    {        if(a[i].va==a[i-1].va&&i!=1)            now[a[i].num]=tot;        else            now[a[i].num]=++tot;    }}void init(){    memset(ans,0,sizeof(ans));    memset(c,0,sizeof(c));    memset(vis,0,sizeof(vis));    memset(now,0,sizeof(now));    for(int i=0;i<=n;i++)        head[i].clear();tot=0;}int main(){int T;scanf("%d",&T);while(T--){scanf("%d",&n);init();for(int i=1;i<n;i++)        {        int a,b;            scanf("%d %d",&a,&b);            add(a,b);        }        for(int i=1;i<=n;i++)        {            int value;            scanf("%d",&value);            a[i].num=i;a[i].va=value;        }        compress();        dfs(1);        for(int i=1;i<=n;i++)            printf("%d\n",ans[i]);}return 0;}

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