hdu5788 level up

最新推荐文章于 2020-06-24 22:45:17 发布

原创最新推荐文章于 2020-06-24 22:45:17 发布 · 525 阅读

0 ·

CC 4.0 BY-SA版权

主席树专栏收录该内容

6 篇文章

订阅专栏

本文详细介绍了一种使用主席树进行区间查询的数据结构实现方法，并结合动态规划解决特定问题。通过具体的代码示例，展示了如何构建主席树、合并节点及查询中位数等关键操作。

贴下以前写的代码
比赛前我准备着重看的
主席树树dp 字符串

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAXN = 1e5+5;
const int MOD = 1e9+7;

int N;
int A[MAXN];
vector<int> mp[MAXN];
int val[MAXN][2];
/**********President Tree**********/
int tot;
struct Node{
    int ls, rs, cc;
    Node(int a=0, int b=0, int c=0):ls(a), rs(b), cc(c){}
}tree[MAXN*20];
int sta[MAXN];
int Build(int pos,int l,int r){
    int rt = tot++;
    tree[rt].cc = 1;
    tree[rt].ls = tree[rt].rs = 0;
    if(l == r) return rt;
    int mid = (l+r) >> 1;
    if(pos <= mid) tree[rt].ls = Build(pos, l,mid);
    else tree[rt].rs = Build(pos, mid+1,r);
    return rt;
}
int Merge(int x,int y,int l,int r){
    if(x == 0 || y == 0) return x+y;
    tree[x].cc += tree[y].cc;
    if(l == r) {
        tree[x].ls = tree[x].rs = 0;
        return x;
    }    
    int mid = (l+r) >> 1;
    tree[x].ls = Merge(tree[x].ls, tree[y].ls, l, mid);
    tree[x].rs = Merge(tree[x].rs, tree[y].rs,mid+1, r);
    return x;
}
int Query(int rt,int K,int l,int r){
    if(tree[rt].cc < K) return 100000;
    if(l == r) return l;
    int mid = (l+r)>>1;
    if(tree[tree[rt].ls].cc >= K) return Query(tree[rt].ls, K, l ,mid);
    else return Query(tree[rt].rs, K-tree[tree[rt].ls].cc, mid+1, r);
}

void dfs(int x){
    sta[x] = Build(A[x], 1, 100000);
    for(int i = 0; i < mp[x].size(); ++i){
        int y = mp[x][i];
        dfs(y);
        sta[x] = Merge(sta[x], sta[y], 1, 100000);
    }
//    printf("%d\n",sta[x]);
    val[x][0] = Query(sta[x], tree[sta[x]].cc+1>>1, 1, 100000);
    val[x][1] = Query(sta[x], (tree[sta[x]].cc+1>>1)+1, 1, 100000);
}
/**************BIT*************/
ll bitt[MAXN];
void Add(int pos,int num){
    for(int i = pos; i > 0; i -= i&-i)
        bitt[i] += num;
}
ll Sum(int pos){
    ll ans = 0;
    for(int i = pos; i <= 100000; i += i&-i)
        ans += bitt[i];
    return ans;
}
ll dfs2(int x){
    Add(val[x][0], val[x][1]-val[x][0]);
    ll ans = Sum(val[x][0]);
    for(int i = 0; i < mp[x].size(); ++i){
        int y = mp[x][i];
        ans = max(ans, dfs2(y));
    }
    Add(val[x][0], val[x][0]-val[x][1]);
    return ans;
}


int main(){
    while(~scanf("%d",&N)){
        memset(bitt,0,sizeof(bitt));

        tot = 1;
        for(int i = 1; i <= N; ++i) scanf("%d",&A[i]);
        for(int i = 1; i <= N; ++i) mp[i].clear();
        for(int i = 2; i <= N; ++i){
            int a; scanf("%d",&a);
            mp[a].push_back(i);
        }
        dfs(1);
        ll sum = 0;
        for(int i = 1; i <= N; ++i) 
            sum += val[i][0];
        printf("%lld\n", sum + dfs2(1));
    }
    return 0;
}