POJ2373 Dividing the Path (monotone queue)

This question is more detail, we first define the state, f [i] represents the processing time to a minimum the number i, in fact, the left edge of the last tube, then these i can not be the place where the cows, because we medals For the left edge

After we discovered monotonicity, in fact, take min + 1, but in the range to be between 2a-2b, so into the queue to be careful when

This time I made the mistake because I regard it as the previous topic and found no restriction here is that the place must be even have to put, because the scope of each tube is even.

#include<iostream>
#include<queue>
#include<map>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<stack>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=1e6+7;
const int inf=0x3f3f3f3f;
int st[N];
struct node{
    int l,r;
}s[N];
int f[N];
int q[N];
int main(){
    int i;
    int n,m;
    int l,a,b;
    cin>>n>>l>>a>>b;
    for(i=1;i<=n;i++){
        cin>>s[i].l>>s[i].r;
        for(int j=s[i].l+1;j<s[i].r;j++)
            st[j]=1;
    }
    memset(f,0x3f,sizeof f);
    f[0]=0;
    int j;
    int hh=0;
    int tt=0;
    q[0]=0;
    for(int i=2*a;i<=l;i+=2){
       while(hh<=tt&&q[hh]<i-2*b)
          hh++;
       while(hh<=tt&&f[q[tt]]>=f[i-2*a])
          tt--;
       q[++tt]=i-2*a;
       if(st[i])
        continue;
       f[i]=f[q[hh]]+1;
    }
    if(f[l]<inf/2)
        cout<<f[l]<<endl;
    else
        cout<<-1<<endl;
}

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