xjtuoj 1264：视频质量
题目分为两个任务，\(X\)是金典的区间全覆盖问题，\(Y\)是woshidashabi
注意区间是左闭右开所以判断取最远覆盖的时候用\(<\)
#include<bits/stdc++.h>
#define F(i,n,m) for(ll i=n;i<m;i++)
#define itn int
#define f(i,n,m) for(ll i=n;i>m;i--)
//#define test
typedef unsigned long long ull;
typedef long long ll;
using namespace std;
struct PP {
ll lp,rp;
} fc[200005];
inline ll read() {
ll num = 0;
char c;
bool flag = false;
while ((c = getchar()) == ' ' || c == '\n' || c == '\r');
if (c == '-') flag = true;
else
num = c - '0';
while (isdigit(c = getchar()))
num = num * 10 + c - '0';
return (flag ? -1 : 1) * num;
}
bool cmp(PP a,PP b) {
if(a.lp==b.lp) return a.rp>b.rp;
return a.lp<b.lp;
}
int main() {
std::ios::sync_with_stdio(false);
ll n,m;
n=read();
m=read();
F(i,0,n) {
fc[i].lp=read();
fc[i].rp=read();
}
sort(fc,fc+n,cmp);
#ifdef test
F(i,0,n) cout<<endl<<fc[i].lp<<" "<<fc[i].rp<<endl;
#endif
ll la,nt;
ll ans=0;
la=nt=0;
F(i,0,n) {
if(la<fc[i].lp) {//左闭右开，找到新的起始点及其最远距离，并且选取上一个起始点计入答案
la=nt;
ans++;
}
nt=max(nt,fc[i].rp);//更新范围内最远距离
if(nt==m) {//如果这次最远距离已经选到边界，直接退出更快
ans++;
break;
}
#ifdef test
cout<<ans<<" "<<i<<" "<<la<<endl;
#endif
}
ll sna=0,san=0;
f(i,n-1,-1) {
if(fc[i].rp==m) sna++;
if(fc[i].lp==0) san++;
}
cout<<ans<<" "<<max(n-san+1,n-sna+1);
return 0;
}
然后是任务2，考虑极端情况即选取的\(Y-1\)个观众都没有看第一或最后一个视频，故答案取\(n-min(num(a_0),num(a_n))+1\)
