[USACO13JAN]Party Invitations【模拟】

本文详细解析了USACO竞赛中Party Invitations问题的解决方案。通过使用vector、queue和set数据结构,存储并处理与每个元素相关的集合。算法描述包括将1号元素加入队列,循环遍历并移除与之相关的集合中的1号元素,检查集合大小。若大小为1,则将该集合元素加入队列,避免重复选择,确保解题的正确性。

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Luogu3068

Sol

这是一篇需要好多容器的题解: vector v e c t o r queue q u e u e set s e t 。我们用 vector v e c t o r 存下与 i i 有关的集合是多少,用set存下每一个集合,用 queue q u e u e 存下被邀请的奶牛。

第一次就是 1 1 入队,然后循环vector,把循环到的集合中的 1 1 都删去,判断删去后的集合大小是否为1,如果是 1 1 ,就入队,重复操作。坑点就是,出来的可能会被重复做,加一个数组判断一下是否已经选了这头奶牛。

Code

#include<iostream>
#include<cstdio>
#include<set>
#include<vector>
#include<queue>
using namespace std;

int n , m , ans , vis[1000005];
set<int>s[250005];
vector<int>about[1000005];
queue<int>q;

int main() {
    scanf("%d%d",&n,&m);
    for(int i=1; i<=m; i++) {
        int t;
        scanf("%d",&t);
        for(int j=1; j<=t; j++) {
            int x;
            scanf("%d",&x);
            about[x].push_back(i);
            s[i].insert(x);
        }
    }
    q.push(1);
    vis[1] = 1;
    while(!q.empty()) {
        int now = q.front();
        q.pop();
        ans++;
        for(int i=0; i<about[now].size(); i++) {
            s[about[now][i]].erase(now);
            if(s[about[now][i]].size() == 1 && !vis[*s[about[now][i]].begin()]) {
                int t = *s[about[now][i]].begin();
                q.push(t);
                vis[t] = 1;
            }
        }
    }
    printf("%d",ans);
    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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