[HAOI2011]problem a
思考一下:转化条件则一个人在 \([a + 1,n - b]\)排名中,且这些人分数都一样。
那么就是若干个区间选择不交集合的问题。
// Problem: P2519 [HAOI2011]problem a
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2519
// Memory Limit: 125 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
#define N 1000005
ll n,cnt,tot;
struct seg{
ll l,r,v;
}e[N],p[N];
ll f[N];
bool operator < (seg a,seg b){
if(a.r != b.r)
return a.r < b.r;
else
return a.l < b.l;
}
#define mid ((l + r) >> 1)
inline ll find(int u){
ll l = 1,r = u - 1;
while(l + 1 < r){
if(p[mid].r < p[u].l)
l = mid;
else
r = mid;
}
while(p[l + 1].r < p[u].l)
++l;
return l;
}
int main(){
scanf("%lld",&n);
for(int i = 1;i <= n;++i){
ll l,r;
scanf("%lld%lld",&l,&r);
ll a = l + 1,b = n - r;
if(a <= b && l + r < n)
e[++cnt].l = a,e[cnt].r = b;
}
std::sort(e + 1,e + cnt + 1);
ll last = 1;
// for(int i = 1;i <= cnt;++i)
// std::cout<<e[i].l<<" "<<e[i].r<<std::endl;
for(int i = 1;i <= cnt + 1;++i){
if(e[last].l != e[i].l || e[last].r != e[i].r)
p[++tot].l = e[last].l,p[tot].r = e[last].r,p[tot].v = i - last,last = i;
}
for(int i = 1;i <= tot;++i)
p[tot].v = std::min(p[tot].v,1ll * p[tot].r - p[tot].l + 1);
// for(int i = 1;i <= tot;++i)
// std::cout<<p[i].l<<" "<<p[i].r<<" "<<p[i].v<<std::endl;
// find(2);
f[1] = p[1].v;
for(int i = 2;i <= tot;++i){
ll k = find(i);
f[i] = std::max(f[i - 1],f[k] + p[i].v);
}
std::cout<<n - f[tot]<<std::endl;
}