hdu 5735 Born Slippy 暴力

最新推荐文章于 2025-08-09 14:59:10 发布

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最新推荐文章于 2025-08-09 14:59:10 发布

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文章标签： php 数据结构与算法

原文链接：http://www.cnblogs.com/qscqesze/p/5692878.html

本文介绍了一道名为 BornSlippy 的 ACM 竞赛题目，并提供了详细的解题思路和代码实现。该题涉及树形结构上的动态规划算法，需要求解每个节点到根路径的最大权值和。

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Born Slippy

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5735

Description

Professor Zhang has a rooted tree, whose vertices are conveniently labeled by 1,2,...,n. And the i-th vertex is assigned with weight wi.

For each s∈{1,2,...,n}, Professor Zhang wants find a sequence of vertices v1,v2,...,vm such that:

1. v1=s and vi is the ancestor of vi−1 (1<i≤m).
2. the value f(s)=wv1+∑i=2mwvi opt wvi−1 is maximum. Operation x opt y denotes bitwise AND, OR or XOR operation of two numbers.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n and a string opt (2≤n≤216,opt∈{AND,OR,XOR}) -- the number of vertices and the operation. The second line contains n integers w1,w2,...,wn (0≤wi<216). The thrid line contain n−1 integers f2,f3,...,fn (1≤fi<i), where fi is the father of vertex i.

There are about 300 test cases and the sum of n in all the test cases is no more than 106.

Output

For each test case, output an integer S=(∑i=1ni⋅f(i)) mod (109+7).

Sample Input

3
5 AND
5 4 3 2 1
1 2 2 4
5 XOR
5 4 3 2 1
1 2 2 4
5 OR
5 4 3 2 1
1 2 2 4

Sample Output

91
139
195

Hint

题意

给你一棵树，树上点有点权，对于每个点，你需要找到到根的那条链上的一个子序列。

使得f[i] = w[v[i]] + sigma w[v[i]] opt w[v[i+1]] 最大。

然后输出sigma(if[i])%mod

题解：

不会正解，n^2dp非常简单，很容易就能想到

然后感觉这道题的数据比较难造，就直接暴力了，然后一发就过了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 70000;
const int mod = 1e9+7;
long long dp[maxn],w[maxn];
int n;
vector<int>E[maxn];
string opt;
set<pair<long long,int> > S;
set<pair<long long,int> >::iterator it;
long long solve(long long x,long long y){
    if(opt=="XOR")return x^y;
    if(opt=="AND")return x&y;
    if(opt=="OR")return x|y;
}
void dfs(int x){
    if(S.size()!=0){
        int tot = 0;
        for(it = S.begin();it!=S.end()&&tot<100;it++,tot++){
            dp[x] = max(dp[x],-(it->first)+solve(w[it->second],w[x]));
        }
    }
    S.insert(make_pair(-dp[x],x));
    for(int i=0;i<E[x].size();i++){
        int v = E[x][i];
        dfs(v);
    }
    S.erase(make_pair(-dp[x],x));
}
void solve(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)E[i].clear(),dp[i]=0;
    cin>>opt;
    for(int i=1;i<=n;i++)
        scanf("%d",&w[i]);
    for(int i=2;i<=n;i++){
        int a;scanf("%d",&a);
        E[a].push_back(i);
    }
    dfs(1);
    long long ans = 0;
    for(int i=1;i<=n;i++){
        ans = (ans+i*(dp[i]%mod+w[i]%mod)) % mod;
    }
    printf("%I64d\n",ans);
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--)solve();
    return 0;
}

转载于:https://www.cnblogs.com/qscqesze/p/5692878.html