Description
N个矩形,排成一排. 现在希望用尽量少的矩形海报Cover住它们.
Input
第一行给出数字N,代表有N个矩形.N在[1,250000] 下面N行,每行给出矩形的长与宽.其值在[1,1000000000]2 1/2 Postering
Output
最少数量的海报数.
Sample Input
5
1 2
1 3
2 2
2 5
1 4
Sample Output
4
HINT
单调递增栈随便搞一下就好
/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline int read(){
int x=0,f=1;char ch=getchar();
for (;ch<\'0\'||ch>\'9\';ch=getchar()) if (ch==\'-\') f=-1;
for (;ch>=\'0\'&&ch<=\'9\';ch=getchar()) x=(x<<1)+(x<<3)+ch-\'0\';
return x*f;
}
inline void print(int x){
if (x>=10) print(x/10);
putchar(x%10+\'0\');
}
const int N=2.5e5;
int stack[N+10],top;
int main(){
int n=read(),same=0;
for (int i=1;i<=n;i++){
int x=read(),y=read();
while (top&&y<=stack[top]){
if (y==stack[top]) same++;
top--;
}
stack[++top]=y;
}
printf("%d\\n",n-same);
return 0;
}