一、PAT A1124 Raffle for Weibo Followers

1. 简单的模拟，细心一点就好啦。参考代码如下。

#include<iostream>
#include<string>
#include<unordered_map>
using namespace std;

unordered_map <string, bool> is_choose;

int main() {
	int n, m, s;
	string name;
	cin >> n >> m >> s;
	bool have = false;
	for (int i = 1, t = s; i <= n; i++) {
		cin >> name;
		if (i == t) {
			if (!is_choose[name]) {
				cout << name << endl;
				t += m;
				is_choose[name] = true;
				have = true;
			}
			else t++;
		}
	}
	if (!have)
		cout << "Keep going..." << endl;
	return 0;
}

二、PAT A1125 Chain the Ropes

1. 显然长绳子的折叠次数越少越好，所以应把长绳子放在最后叠，所以此题即为堆果子问题，利用优先级队列即可解决。注意最后的输出。参考代码如下。

#include<cstdio>
#include<vector>
#include<queue>
#include<math.h>
using namespace std;

priority_queue <float, vector<float>, greater<float>> pq;

int main() {
	int n;
	float t, a, b;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%f", &t);
		pq.push(t);
	}
	for (int i = 1; i < n; i++) {
		a = pq.top(); pq.pop();
		b = pq.top(); pq.pop();
		pq.push((a + b) / 2.0);
	}
	printf("%.0f\n", floor(pq.top()));
	return 0;
}

三、PAT A1126 Eulerian Path

1. 首先需要判断图是否连通，这里用 dfs 进行一次遍历，如果有未访问到的点说明图不连通。如果是连通图，则要考虑顶点的度的奇偶。如果恰有两个点的度为奇数，则是 Semi-Eulerian，如果顶点的度数全为偶数则是 Eulerian，否则即为 Non-Eulerian。参考代码如下。

#include<iostream>
#include<vector>
using namespace std;
const int maxn = 505;

vector <int> adj[maxn];
bool vis[maxn];
void dfs(int r) {
	vis[r] = true;
	for (auto it : adj[r])
		if (!vis[it]) dfs(it);
}

int main() {
	int n, m, u, v, cnt = 0;
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1);
	for (int i = 1; i <= n; i++) {
		if (i != 1) cout << ' ';
		cout << adj[i].size();
		if (adj[i].size() % 2 == 1) cnt++;
	}
	cout << endl;
	for (int i = 1; i <= n; i++)
		if (!vis[i]) {
			cout << "Non-Eulerian" << endl;
			return 0;
		}
	if (cnt == 0) cout << "Eulerian" << endl;
	else if (cnt == 2) cout << "Semi-Eulerian" << endl;
	else cout << "Non-Eulerian" << endl;
	return 0;
}

四、PAT A1127 ZigZagging on a Tree

1. 先建树，然后层次遍历，同时记录每个结点的所在层次，存储层次遍历的结果。输出时将层次为偶数的一连串结点正序输出，将层次为奇数的一连串结点逆序输出，逆序输出时使用了栈。参考代码如下。

#include<iostream>
#include<vector>
#include<queue>
#include<stack>

using namespace std;

struct node {
	int v;
	int level;
	node* left = NULL;
	node* right = NULL;
};
int in[50], post[50];
vector <node*> ans;
 
node* build(int inL, int inR, int postL, int postR) {
	if (inL > inR) return NULL;
	node* root = new node;
	root->v = post[postR];
	int k;
	for (k = inR; k >= inL; k--)
		if (in[k] == post[postR]) break;
	int num = inR - k;
	root->left = build(inL, k - 1, postL, postR - num - 1);
	root->right = build(k + 1, inR, postR - num, postR - 1);
	return root;
}
void level(node* root) {
	queue <node*> q;
	root->level = 1;
	q.push(root);
	ans.push_back(root);
	while (q.size()) {
		node* temp = q.front();
		q.pop();
		if (temp->left) {
			temp->left->level = temp->level + 1;
			q.push(temp->left);
			ans.push_back(temp->left);
		}
		if (temp->right) {
			temp->right->level = temp->level + 1;
			q.push(temp->right);
			ans.push_back(temp->right);
		}
	}
}

int main() {
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) cin >> in[i];
	for (int i = 0; i < n; i++) cin >> post[i];
	node* root = build(0, n - 1, 0, n - 1);
	level(root);
	cout << ans[0]->v;
	int i = 1;
	while (i < ans.size()) {
		if (ans[i]->level % 2 == 0)
			cout << ' ' << ans[i++]->v;
		else {
			stack <node*> s;
			while (i < ans.size() && ans[i]->level % 2 == 1)
				s.push(ans[i++]);
			while (s.size()) {
				cout << ' ' << s.top()->v;
				s.pop();
			}
		}
	}
	return 0;
}

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来源： https://blog.csdn.net/weixin_42717165/article/details/88052651