Description

給你一個n長度的數軸和m個區間，每個區間裏有且僅有一個點，問能有多少個點

Input

* Line 1: Two integers N and M.

* Lines 2..M+1: Line i+1 contains a_i and b_i.

Output

* Line 1: The maximum possible number of spotted cows on FJ‘s farm, or -1 if there is no possible solution.

Sample Input

Sample Output

#include<cstdio>
#include<cstring>
#include<algorithm>
using std::min;
using std::max;
const int M=250007,inf=0x3f3f3f3f;
int read(){
    int ans=0,f=1,c=getchar();
    while(c<‘0‘||c>‘
 
9‘){if(c==‘-‘) f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){ans=ans*10+(c-‘0‘); c=getchar();}
    return ans*f;
}
int n,m,x,y;
int l[M],r[M],f[M];
int q[M],ql=1,qr;
int main(){
    n=read(); m=read();
    for(int i=1;i<=n+1;i++) r[i]=i-1;
    for(int i=1;i<=m;i++){
        x=read(); y=read();
        r[y]=min(r[y],x-1);
        l[y+1]=max(l[y+1],x);
    }
    for(int i=n;i;i--) r[i]=min(r[i],r[i+1]);
    for(int i=2;i<=n+1;i++) l[i]=max(l[i],l[i-1]);
    f[qr=1]=0;
    for(int i=1;i<=n+1;i++){
        for(int k=r[i-1]+1;k<=r[i];k++){
            while(ql<=qr&&f[q[qr]]<=f[k]) qr--;
            q[++qr]=k;
        }
        while(ql<=qr&&q[ql]<l[i]) ql++;
        if(ql>qr) f[i]=-inf;
        else f[i]=f[q[ql]]+1;
    }
    if(f[n+1]>=0) printf("%d\n",f[n+1]-1);
    else printf("-1\n");
    return 0;
}

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