Each question is intended for a sprinkler radius [A, B], each cow has a section of its own [s, e], this range can be covered by a sprinkler.
Requires the entire interval [1, L] (L <= 1e6) is covered does not overlap, at least to the number of sprinklers, sprinklers should not exceed the range of the overall range interval.
Since an interval [s, e] only covered by a sprinkler, so [s + 1, e-1] can not be a certain range of watering sprinkler right endpoint.
We not_r [] to make a mark.
L to see the range of 1e6, we should know is O (n) algorithm to use.
Set f [i] represents the number of sprinklers i to the desired position. (I must be the right end point ).
f[i]=min（f[j]）+1（A<=（i-j）/2<=B）
Direct cycle time O (n ^ 2) optimization considerations.
We found that for point i, j must be used for a continuous range, so a single team to maintain a minimum of f [j].
Since a j can be used depends on its location meets limited, id stored at the queue with the subscript.
Because i was right point, so when the cycle, we are an even number increases.
#include<cstdio> #include<iostream> #include<cstring> #define LL long long #define INF 2100000000 #define N 1000003 #define re register using namespace std; int read() { int x=0,f=1;char s=getchar(); while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();} while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();} return x*f; } void print(int x) { if(x<0)x=-x,putchar('-'); if(x>9)print(x/10); putchar(x%10+'0'); } int id[N],f[N],q[N],not_r[N]; int main() { int n=read(),l=read(); int a=read(),b=read(); for(int i=1;i<=n;++i) { int s=read(),t=read(); for(int j=s+1;j<t;++j) not_r[j]=1; } for(int i=1;i<=l;++i)f[i]=INF; int h=1,t=0,head=1,tail=0; id[++t]=0; for(int i=2;i<=l;i+=2)//偶数个的加 { if(not_r[i])continue; while(h<=t&&(i-id[h])/2>=a) { while(head<=tail&&f[q[tail]]>=f[id[h]])tail--; q[++tail]=id[h];//加入q里了，就可以移出id了 h++; } while(head<=tail&&(i-q[head])/2>b)head++; if(head<=tail) { f[i]=f[q[head]]+1; id[++t]=i; } } if(f[l]!=INF)printf("%d\n",f[l]); else printf("-1\n"); } /* */