[POI2008]PLA-Postering

题目链接

[POI2008]PLA-Postering

description

春春幼儿园升级版

solution

此题数据较强,于是我们可以用单调栈来维护.具体维护方式如下:

首先将\(ans\)初始化为\(n\),对于当前高度\(h\),如果当前栈顶大于\(h\),则将栈顶弹出,直到栈顶不大于\(h\)或栈空.进行此番操作后,将栈顶元素与\(h\)比较,若相等则\(--ans\).然后将当前元素入栈即可.

正确性证明:(大眼观察法)我们容易发现,宽度是无关变量,答案只与高度有关.而且,\(n\)是肯定满足答案的,所以我们可以从\(n\)开始考虑.欲省掉一张海报,当且仅当两值之间的值均大于此二值且此二值相等,所以可以用单调栈维护.

code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#define R register
#define next jkflsdlkf
#define debug puts("mlg")
#define mod 10007
#define Mod(x) ((x%mod+mod)%mod)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
inline ll read();
inline void write(ll x);
inline void writesp(ll x);
inline void writeln(ll x);
ll n,_stack[1000000],t,ans;
int main(){
	ans=n=read();
	for(R ll i=1,h;i<=n;i++){
		read();h=read();
		while(t&&_stack[t]>h) t--;
		if(_stack[t]==h) --ans;
		_stack[++t]=h;
	}
	writeln(ans);
} 
inline ll read(){ll x=0,t=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') t=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*t;}
inline void write(ll x){if(x<0){putchar('-');x=-x;}if(x<=9){putchar(x+'0');return;}write(x/10);putchar(x%10+'0');}
inline void writesp(ll x){write(x);putchar(' ');}
inline void writeln(ll x){write(x);putchar('\n');}