USACO 1月 2021-2022 January Contest Silver银组 题解
你好啊我又又又来了
要准备usaco的铁铁们可以参考这个文章哦！
想刷好USACO——看这篇文章就够了_GeekAlice的博客-CSDN博客我最近是发现了一个很好用的网站
https://blog.csdn.net/GeekAlice/article/details/122291933?spm=1001.2014.3001.5501
这次是银组的题解，
如果你想看铜组的话，传送门在这里：
USACO 1月 2021-2022 Contest Bronze 题解_GeekAlice的博客-CSDN博客USACO 2021-22 January Contest Bronze 完整版题解https://blog.csdn.net/GeekAlice/article/details/122798286?spm=1001.2014.3001.5502
:))
话不多说，上题解：
✓
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版权声明：本文为CSDN博主「GeekAlice」的原创文章，遵循CC 4.0 BY-SA版权协议，转载请附上原文出处链接及本声明。
题解：
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;public class SearchingForSoulmates {public static void main(String[] args) throws IOException {BufferedReader in = new BufferedReader(new InputStreamReader(System.in));StringBuilder out = new StringBuilder();for (int t = Integer.parseInt(in.readLine()); t > 0; t--) {StringTokenizer tokenizer = new StringTokenizer(in.readLine());long cow1 = Long.parseLong(tokenizer.nextToken());long cow2 = Long.parseLong(tokenizer.nextToken());long answer = Long.MAX_VALUE;for (int removed = 0; cow2 >> removed > 0; removed++) {long here = 0;long prefix = cow2 >> removed;long cow = cow1;while (cow > prefix) {if (cow % 2L == 1L) {cow++;here++;}cow /= 2L;here++;}here += prefix - cow;here += removed;here += Long.bitCount(cow2 & ((1L << removed) - 1L));answer = Math.min(answer, here);}out.append(answer).append('\n');}System.out.print(out);}
}
#include <bits/stdc++.h>
using namespace std;
int64_t ans = 0;
int N;
void add_contribution(const vector<int>& h) {vector<int> with_h(N+1);for (int i = 0; i < N; ++i) with_h[h[i]] = i;set<int> present;for (int cur_h = N; cur_h; --cur_h) {auto it = present.insert(with_h[cur_h]).first;if (next(it) != end(present)) ans += *next(it)-*it+1;}
}
void add_contribution_ll(const vector<int>& h) {vector<int> with_h(N+1);for (int i = 0; i < N; ++i) with_h[h[i]] = i;vector<int> pre(N), nex(N);for (int i = 0; i < N; ++i) {pre[i] = i-1;nex[i] = i+1;}for (int cur_h = 1; cur_h <= N; ++cur_h) {int pos = with_h[cur_h];int p = pre[pos], n = nex[pos];if (n != N) ans += n-pos+1, pre[n] = p;if (p != -1) nex[p] = n;}
}
void add_contribution_alt(const vector<int>& h) {stack<int> stk;for (int i = N-1; i >= 0; --i) {while (!stk.empty() && h[stk.top()] < h[i]) stk.pop();if (!stk.empty()) ans += stk.top()-i+1;stk.push(i);}
}int main() { cin >> N;vector<int> h(N); for (int& i: h) cin >> i;add_contribution(h);reverse(begin(h), end(h));add_contribution(h);cout << ans;
}
#include <bits/stdc++.h>
using namespace std;
struct edge {int cow; int to;bool is_first;edge() {};edge(int cow, int to, bool is_first) : cow(cow), to(to), is_first(is_first) {};
};
vector<edge> adj[100001];
bool visited_cycle[100001];
bool visited[100001];
bool gets_cereal[100001];
int hungry_cows = 0;
queue<int> order;
int ignore_edge = -1,N,M,first_cereal = -1; void find_cycle(int cur, int prev = -1) {visited_cycle[cur] = true; for (edge next : adj[cur]) {if (visited_cycle[next.to]) {if (first_cereal == -1 && next.to != prev) {if (next.is_first) {first_cereal = next.to; } else {first_cereal = cur;}ignore_edge = next.cow; order.push(next.cow);gets_cereal[next.cow] = true;}} else {find_cycle(next.to, cur);}}
}void dfs(int cur) {visited[cur] = true;for (edge next : adj[cur]) {if (!visited[next.to] && next.cow != ignore_edge) { gets_cereal[next.cow] = true;order.push(next.cow);dfs(next.to);}}
}int main() {cin >> N >> M;for (int i = 0; i < N; ++i) {int a, b;cin >> a >> b;adj[a].push_back(edge(i+1, b, false));adj[b].push_back(edge(i+1, a, true));}for (int i = 1; i <= M; ++i) {first_cereal = -1;ignore_edge = -1;if (!visited[i]) {find_cycle(i);if (first_cereal > 0) {dfs(first_cereal);} else {dfs(i);}}}for (int i = 1; i <= N; ++i) {if (!gets_cereal[i]) {++hungry_cows;order.push(i);} }cout << hungry_cows << endl;while (!order.empty()) {cout << order.front() << endl; order.pop();}return 0;
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