AtCoder Regular Contest 076 F - Exhausted? （贪心）

Topic

answer

If only llThe limit of$l$ , then it must be directly in accordance with$l$ Sort from small to large, and then fill in from the chair on the left one by one. It is a greedy board question.

But this question also limits $l$ andrrWhat about$r$ , it has directly evolved into a black question.

We might as well follow $l$ Greed, then give$r$ An opportunity to repent.

A bit abstract, the specific approach is:

We still follow $l$ Sort (from small to large, as mentioned above), each time the smallest one is not occupied, if there is no such position, then consider "replace" the previous person, and therrof the person who needs to be replaced$r\; is$ higher than the current$r\; is$ small (so that it can be replaced with more options), if there are more than one that meets the requirements,rrshould be replaced obviously$The$ smallest$r$ comes out.
After the operation, there are still a few people left (unlucky to be "replaced" or "replaced" failed), and the occupied position must be a prefix at this time, so we press (the remaining people)$r$ Sort from the largest to the smallest and then be greedy (start from the chair on the right). At this time, people who can’t sit down have no choice.

Seeing the "obvious" above, you know that it is definitely not what the author said. The above method is from the comment of the wzy god, and it feels very understandable (I only explained four places).

So only need to do llUse a small root heap to maintainrrwhen$l$$r$ smallest people, and secondly to pay attention to detail, behind greedyrrCan't sit in front of greedyllwhen$r$Seated position at$l$ .

CODE

#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define MAXN 200005
#define LL long long
#define DB double
#define ENDL putchar('\n')
LL read() {
    
    
	LL f=1,x=0;char s = getchar();
	while(s < '0' || s > '9') {
    
    if(s=='-')f = -f;s = getchar();}
	while(s >= '0' && s <= '9') {
    
    x=x*10+(s-'0');s = getchar();}
	return f * x;
}
int n,m,i,j,s,o,k;
struct it{
    
    
	int l,r;  it(){
    
    l=r=0;}
	it(int L,int R){
    
    l=L;r=R;}
}a[MAXN];
bool cmp(it a,it b) {
    
    return a.l < b.l;}
bool operator < (it a,it b) {
    
    return a.r < b.r;}
bool operator > (it a,it b) {
    
    return b < a;}
priority_queue<it,vector<it>,greater<it> > b1;
priority_queue<it> b2;
int main() {
    
    
	n = read();m = read();
	for(int i = 1;i <= n;i ++) {
    
    
		a[i].l = read();a[i].r = read();
	}
	sort(a + 1,a + 1 + n,cmp);
	int le = 0;
	b1.push(it(0,m+1));
	for(int i = 1;i <= n;i ++) {
    
    
		if(le < a[i].l) {
    
    
			le ++;
			b1.push(a[i]);
		}
		else {
    
    
			if(a[i].r > b1.top().r) {
    
    
				b2.push(b1.top());
				b1.pop(); b1.push(a[i]);
			}
			else b2.push(a[i]);
		}
	}
	int ans = 0,le2 = m+1;
	while(!b2.empty()) {
    
    
		it t = b2.top(); b2.pop();
		if(le2 > max(t.r,le+1)) le2 --; // Detail!! : Don't disturb the prefix !!
		else ans ++;
	}
	printf("%d\n",ans);
	return 0;
}