UVA 112(Tree Summing)回溯

本文介绍了一种解决树形结构中从根节点到叶子节点路径权值总和与给定值比较的问题。通过递归回溯的方法实现,判断路径权值之和是否等于预设值,最终输出yes或no。

题意:给出一个sum值,判断一棵树沿着根节点到叶子节点路径上所有节点权值之和与sum是否相等

回溯法!自己写的算法疯狂TLE,看了大佬的博客,誓与大佬同归于尽

#include<bits/stdc++.h>
#define LL long long
#define maxn 100010
using namespace std;
int sum;
bool flag=false;
bool tree_sum(int n)
{
    int num;
    char ch;
    cin>>ch;
    if(cin>>num)
    {
        n+=num;
        bool temp=tree_sum(n)|tree_sum(n);
        if(!temp&&sum==n)
        {
            flag=true;
        }
        cin>>ch;
        return true;
    }
    else
    {
        cin.clear();
        cin>>ch;
        return false;
    }
}
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    while(cin>>sum)
    {
        flag=false;
        tree_sum(0);
        if(flag)
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}


### USACO 2016 January Contest Subsequences Summing to Sevens Problem Solution and Explanation In this problem from the USACO contest, one is tasked with finding the size of the largest contiguous subsequence where the sum of elements (IDs) within that subsequence is divisible by seven. The input consists of an array representing cow IDs, and the goal is to determine how many cows are part of the longest sequence meeting these criteria; if no valid sequences exist, zero should be returned. To solve this challenge efficiently without checking all possible subsequences explicitly—which would lead to poor performance—a more sophisticated approach using prefix sums modulo 7 can be applied[^1]. By maintaining a record of seen remainders when dividing cumulative totals up until each point in the list by 7 along with their earliest occurrence index, it becomes feasible to identify qualifying segments quickly whenever another instance of any remainder reappears later on during iteration through the dataset[^2]. For implementation purposes: - Initialize variables `max_length` set initially at 0 for tracking maximum length found so far. - Use dictionary or similar structure named `remainder_positions`, starting off only knowing position `-1` maps to remainder `0`. - Iterate over given numbers while updating current_sum % 7 as you go. - Check whether updated value already exists inside your tracker (`remainder_positions`). If yes, compare distance between now versus stored location against max_length variable's content—update accordingly if greater than previous best result noted down previously. - Finally add entry into mapping table linking latest encountered modulus outcome back towards its corresponding spot within enumeration process just completed successfully after loop ends normally. Below shows Python code implementing described logic effectively handling edge cases gracefully too: ```python def find_largest_subsequence_divisible_by_seven(cow_ids): max_length = 0 remainder_positions = {0: -1} current_sum = 0 for i, id_value in enumerate(cow_ids): current_sum += id_value mod_result = current_sum % 7 if mod_result not in remainder_positions: remainder_positions[mod_result] = i else: start_index = remainder_positions[mod_result] segment_size = i - start_index if segment_size > max_length: max_length = segment_size return max_length ```
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