1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #define M 50008 5 using namespace std; 6 struct shu 7 { 8 int l,r,ma; 9 }shu[4*M]; 10 int l,r,n,m,b[M],a1[M],a2[M]; 11 void jian(int b1,int l,int r) 12 { 13 shu[b1].l=l; 14 shu[b1].r=r; 15 if(l==r) 16 { 17 shu[b1].ma=a2[l]; 18 return; 19 } 20 int mid=(l+r)>>1; 21 jian(b1*2,l,mid); 22 jian(b1*2+1,mid+1,r); 23 shu[b1].ma=max(shu[b1*2].ma,shu[b1*2+1].ma); 24 return; 25 } 26 int zhao(int b1,int l,int r) 27 { 28 if(shu[b1].l>=l&&shu[b1].r<=r) 29 return shu[b1].ma; 30 int mid=(shu[b1].l+shu[b1].r)>>1,mx=-100000000; 31 if(l<=mid) 32 mx=max(zhao(b1*2,l,r),mx); 33 if(r>mid) 34 mx=max(zhao(b1*2+1,l,r),mx); 35 return mx; 36 } 37 int main() 38 { 39 scanf("%d",&n); 40 for(int i=1;i<=n;i++) 41 scanf("%d%d",&a1[i],&a2[i]); 42 jian(1,1,n); 43 scanf("%d",&m); 44 for(int i=1;i<=m;i++) 45 { 46 int b1,b2,b3,c1,c2; 47 scanf("%d%d",&c1,&c2); 48 b1=lower_bound(a1+1,a1+n+1,c1)-a1; 49 b2=lower_bound(a1+1,a1+n+1,c2)-a1; 50 if(a1[b1]==c1) 51 b3=zhao(1,b1+1,b2-1); 52 else 53 b3=zhao(1,b1,b2-1); 54 if(a1[b1]==c1&&a1[b2]==c2&&a2[b1]>a2[b2]&&a2[b2]>b3&&b2-1-b1==c2-c1-1) 55 printf("true "); 56 else if((a1[b1]==c1&&a1[b2]!=c2&&a2[b1]<=b3)||(a1[b2]==c2&&a1[b1]!=c1&&a2[b2]<=b3) 57 ||(a1[b1]==c1&&a1[b2]==c2&&(a2[b1]<a2[b2]||a2[b2]<=b3))) 58 printf("false "); 59 else 60 printf("maybe "); 61 } 62 return 0; 63 }
显然的线段树，然而判断条件有点复杂
x，y都有且满足条件，且他们中间数都有为true
否则 一旦有一个x或y存在不满足条件 为false
否则 为maybe