「CEOI2019」魔法树

description

solution

设$d{p}_{i,j}:i$子树在$j$时刻的最大果汁量，显然$d{p}_{i,j}$在$j$是单调递增的 $d{p}_{i,j}=\max (d{p}_{i,j},d{p}_{i,j-1})$

• $i$不收获

  $d{p}_{i,j}={\sum }_{k\in so{n}_i}d{p}_{k,j}$

• $i$在$j$时刻收获，要保证$j\ge d_i$

  $d{p}_{i,j}=w_i+{\sum }_{k\in so{n}_i}d{p}_{k,d_i}$

只有$n$个时刻，可以离散化，暴力转移是$O(n^2)$

可以考虑差分map进行启发式合并

在$i$成熟的$t$时刻时，进行$d{p}_{i,t}+=w_i$

此时会出现$j>t$但是$d{p}_{i,j}<d{p}_{i,t}$的情况，直接删除即可

code

#include <map>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std; 
#define int long long
#define maxn 100005
map < int, int > dp[maxn];
vector < int > G[maxn], fruit[maxn];
int n, m, k;
int d[maxn], w[maxn];bool cmp( int x, int y ) {return d[x] == d[y] ? w[x] < w[y] : d[x] < d[y];
}void dfs( int now ) {for( int son : G[now] ) {dfs( son );if( dp[now].size() < dp[son].size() )swap( dp[now], dp[son] );for( auto i = dp[son].begin();i != dp[son].end();i ++ )dp[now][i -> first] += i -> second;dp[son].clear();}sort( fruit[now].begin(), fruit[now].begin(), cmp );for( int son : fruit[now] ) {dp[now][d[son]] += w[son]; int r = w[son];for( auto i = dp[now].upper_bound( d[son] ), t = i;i != dp[now].end(); ) {if( r >= i -> second ) {r -= i -> second;t = i ++;dp[now].erase( t );}else {i -> second -= r;break;}}}
}signed main() {scanf( "%lld %lld %lld", &n, &m, &k );for( int i = 2, fa;i <= n;i ++ ) {scanf( "%lld", &fa );G[fa].push_back( i );}for( int i = 1, v;i <= m;i ++ ) {scanf( "%lld %lld %lld", &v, &d[i], &w[i] );fruit[v].push_back( i );}dfs( 1 );int ans = 0;for( auto i = dp[1].begin();i != dp[1].end();i ++ )ans += i -> second;printf( "%lld\n", ans );return 0;
}