P3029 [USACO11NOV] Cow Lineup S（双指针）

这道题题标签虽然标了一大堆，但我感觉我只用到了双指针，首先set容器储存id有多少种，map储存id有多少种，当map.size==set.size说明该区间包含了所有种类的奶牛，让c【i】.x-c[j].x;
然后让j对应的id数量--；如果map.size还等于set.size说明可以继续操作

#include<iostream>
#include<set>
#include<map>
#include<algorithm>
#include<vector>
#include<cmath>
#include<climits>
#include<cstring>
#define int long long
const int N = 1e6+5;
using namespace std;
char* p1, * p2, buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{
	int x = 0, f = 1;
	char ch = nc();
	while (ch < 48 || ch>57)
	{
		if (ch == '-')
			f = -1;
		ch = nc();
	}
	while (ch >= 48 && ch <= 57)
		x = x * 10 + ch - 48, ch = nc();
	return x * f;
}
struct cow {
	int x;
	int id;
}c[N];
bool cmp(cow c1, cow c2) {
	return c1.x < c2.x;
}
int x[N], id[N];
int ans = 1e12;
map<int, int>m;
set<int>s;
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin >> n;
	int j = 1;
	for (int i = 1; i <= n; i++) {
		cin >> c[i].x >> c[i].id;
		s.insert(c[i].id);
	}
	sort(c + 1, c + 1 + n,cmp);
	for (int i = 1; i <= n; i++) {
		auto it = m.find(c[i].id);
		if (it == m.end())m.insert(pair<int, int>(c[i].id, 1));
		else it->second++;
		while (m.size() == s.size()) {
			ans = min(ans, c[i].x - c[j].x);
			auto it = m.find(c[j].id);
			it->second--;
			if (it->second == 0)m.erase(c[j].id);
			j++;
		}
	}
	cout << ans;
	return 0;
}