11/10训练笔记

P7831[CCO2021] Travelling Merchant 题解

考虑出度为0的点显然不行

随后，进行一个类似于拓扑排序的过程即可

注意到需要建反图

原图也得保留

注意判-1

代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
struct edge{
	int a,b,r,p,id;
	edge(int a = 0,int b = 0,int r = 0,int p = 0,int id = 0):a(a),b(b),r(r),p(p),id(id){}
	bool operator<(edge o) {
		return r > o.r;
	}
}E1[200010];
int ans[200010],vis[200010],out[200010],n,m,cnt;
vector<pair<pair<int,int>,pair<int,int> > > E[200010];
queue<int> q;
int main()
{
	cin >> n >> m;
	for(int i = 1;i <= m;i++) {
		cin >> E1[i].a >> E1[i].b >> E1[i].r >> E1[i].p;
		E1[i].id = i;
		E[E1[i].b].push_back({{E1[i].a,E1[i].id},{E1[i].r,E1[i].p}});
		out[E1[i].a]++;
	}
	for(int i = 1;i <= n;i++) {
		if(!out[i]) q.push(i);
	}
	sort(E1 + 1,E1 + m + 1);
	memset(ans,0x3f,sizeof(ans));
	for(int i = 1;i <= m;i++) {
		while(!q.empty()) {
			int u = q.front();
			q.pop();
			//cout << u << " ";
			for(auto j:E[u]) {
				int v = j.first.first,id = j.first.second,r = j.second.first,p = j.second.second;
				if(vis[id]) continue;
				if(ans[u] != 0x3f3f3f3f) ans[v] = min(ans[v],max(r,ans[u] - p));
				vis[id] = true;
				out[v]--;
				if(!out[v]) {
					q.push(v);
				}
				//cout << v << " ";
			}
		}
		if(!vis[E1[i].id]) {
			vis[E1[i].id] = true;
			ans[E1[i].a] = min(ans[E1[i].a],E1[i].r);
			out[E1[i].a]--;
			if(!out[E1[i].a]) q.push(E1[i].a);
		}
	}
	for(int i = 1;i <= n;i++) {
		cout << (ans[i] == 0x3f3f3f3f ? -1 : ans[i]) << " ";
	}
	cout << "\n";
}

P7261[COCI2009-2010#3] PATULJCI 题解

首先对每个颜色开一个vector<int>保存其位置，随后对于一段区间\([l,r]\)和一个颜色\(c\)，可以很快速的求出\([l,r]\)内\(c\)出现的次数。

然后进行随机化，每次随机一个元素并查看他的出现次数。

若随机\(t\)次，那么随机不到的概率是\(\frac{1}{2 ^ t}\)，当\(t\)取50时约等于\(10 ^ {-15}\)，可以认为几乎不可能。

代码：

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int t = 50;
int a1[300010],n,c,m,a,b;
vector<int> E[30010];
int query(int l,int r) {
	for(int i = 1;i <= t;i++) {
		int p = l + rand() % (r - l + 1);
		int res = upper_bound(E[a1[p]].begin(),E[a1[p]].end(),r) - lower_bound(E[a1[p]].begin(),E[a1[p]].end(),l);
		if(res > (r - l + 1) / 2) {
			return a1[p];
		}
	}
	return -1;
}
int main()
{
	cin >> n >> c;
	for(int i = 1;i <= n;i++) {
		cin >> a1[i];
		E[a1[i]].push_back(i);
	}
	for(int i = 1;i <= c;i++) {
		E[i].push_back(n + 1);
	}
	cin >> m;
	for(int i = 1;i <= m;i++) {
		cin >> a >> b;
		int ans = query(a,b);
		if(~ans) cout << "yes" << " " << ans << "\n";
		else cout << "no" << "\n";
	}
}

P6706[COCI2010-2011#7] KUGLICE 题解

首先离线，删除转添加

然后并查集维护，其中如果出现环，那么把两个定位到一个虚拟节点上即可

代码：

#include<iostream>
using namespace std;
int out[300010],vis[300010],fa[300010],ans[300010],n,q,opt,x,cnt;
pair<int,int> qu[300010];
void init() {
	for(int i = 1;i <= n + 1;i++) fa[i] = i;
}
int find(int x) {
	return x == fa[x] ? x : fa[x] = find(fa[x]);
}
void merge(int x,int y) {
	int fx = find(x),fy = find(y);
	if(fx == fy) {
		fa[fx] = fa[fy] = n + 1;
	} else {
		fa[fx] = fy;
	}
}
int main()
{
	cin >> n;
	init();
	for(int i = 1;i <= n;i++) {
		cin >> out[i];
	}
	cin >> q;
	for(int i = 1;i <= q;i++) {
		cin >> opt >> x;
		qu[i] = {opt,x};
		if(opt == 2) {
			vis[x] = 1;
		}
	}
	for(int i = 1;i <= n;i++) {
		if(!vis[i] && out[i] != 0) {
			merge(i,out[i]);
		}
	}
	for(int i = q;i >= 1;i--) {
		int opt = qu[i].first,x = qu[i].second;
		if(opt == 2) {
			merge(x,out[x]);
		} else {
			int ans1 = find(x);
			ans[++cnt] = ans1;
		}
	}
	for(int i = cnt;i >= 1;i--) {
		if(ans[i] == n + 1) {
			cout << "CIKLUS" << "\n";
		} else cout << ans[i] << "\n";
	}
	cout << "\n";
}