bzoj 3126 单调队列优化dp

能转移的最左是其左边完整区间的最右左端点,最右是能覆盖它的最左左端点-1

#pragma GCC optimize ("O3")
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 200005
using namespace std;
int l[N],r[N],n,m,f[N],q[N];
int read(){
	int a=0;char ch=getchar();
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch>='0'&&ch<='9'){a=a*10+(ch^48);ch=getchar();}
	return a;
}
int main(){
	n=read(); m=read();
	for(register int i=1;i<=n+1;++i)r[i]=i-1;
	for(register int i=1,x,y;i<=m;++i){
		x=read(); y=read();
		r[y]=min(r[y],x-1);
		l[y+1]=max(l[y+1],x);
	}
	for(register int i=2;i<=n+1;++i)
		l[i]=max(l[i],l[i-1]);
	for(register int i=n;i>=1;--i)
		r[i]=min(r[i],r[i+1]);
	/*for(int i=1;i<=n;i++)
		printf("%d  %d  %d\n",i,l[i],r[i]);*/
	for(register int i=1,j=1,head=1,tail=1;i<=n+1;++i){
		while(j<=r[i]&&j<=n){
			if(f[j]==-1){j++;continue;}
			while(head<=tail&&f[j]>f[q[tail]])--tail;
			q[++tail]=j++;
		}
		while(head<=tail&&q[head]<l[i])++head;
		if(head<=tail)f[i]=f[q[head]]+1;
		else f[i]=-1;
		//printf("%d  %d\n",i,f[i]);
	}
	if(f[n+1]==-1)printf("%d\n",f[n+1]);
	else printf("%d\n",f[n+1]-1);
	return 0;
}


posted @ 2017-09-24 07:49  Ren_Ivan  阅读(105)  评论(0编辑  收藏  举报