Interval modification Point Query HDU1556

1#include <iostream>2#include <cstdio>3 4 using namespacestd;5 6 structNode7 {8     intL,r;9     intv;Ten     intLZ; One}bn[400000]; A  - voidBuildintKintLintR) - { theBn[k].l=l; -Bn[k].r=R; -bn[k].lz=0; -     if(bn[k].l==BN[K].R) +     { -bn[k].v=0; +         return ; A     } at     intlk=k*2; -     intrk=lk+1; -     intMid= (L+R)/2; - build (Lk,l,mid); -Build (rk,mid+1, R); -bn[k].v=0; in } -  to voidPushintk) + { -     intlk=k*2; the     intrk=lk+1; *bn[lk].v+=Bn[k].lz; $bn[rk].v+=Bn[k].lz;Panax Notoginsengbn[lk].lz+=Bn[k].lz; -bn[rk].lz+=Bn[k].lz; thebn[k].lz=0; + } A  the voidChangeintKintLintR) + { -     if(bn[k].l==l&&bn[k].r==R) $     { $bn[k].lz++; -bn[k].v=bn[k].lz* (bn[k].r-bn[k].l+1); -         return ; the     } -     if(BN[K].LZ)Wuyi push (k); the     intlk=k*2; -     intrk=lk+1; Wu     if(bn[lk].r>=R) - Change (lk,l,r); About     Else if(bn[rk].l<=l) $ Change (rk,l,r); -     Else -     { - Change (LK,L,BN[LK].R); A Change (rk,bn[rk].l,r); +     } thebn[k].v=bn[rk].v+bn[lk].v; - } $  the intans; the voidSearchintKinti) the { the     if(bn[k].l==i&&bn[k].r==i) -     { inans=bn[k].v; the         return ; the     } About     if(BN[K].LZ) the push (k); the     intlk=k*2; the     intrk=lk+1; +     if(bn[lk].r>=i) - search (lk,i); the     Else if(bn[rk].l<=i)Bayi search (rk,i); thebn[k].v=bn[rk].v+bn[lk].v; the } -  - intMain () the { the     intN; the      while(cin>>n&&N) the     { -         intb; theBuild1,1, n); the          for(intI=1; i<=n;i++) the         {94scanf"%d%d",&a,&b); theChange1, A, b); the         } theSearch1,1);98cout<<ans; About          for(intI=2; i<=n;i++) -         {101Search1, i);102cout<<" "<<ans;103         }104cout<<Endl; the     }106     return 0;107}

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This is normal line segment tree practice

1#include <iostream>2#include <cstdio>3#include <cstring>4 5 using namespacestd;6 7 intan[100010];8 intans[100010];9 Ten intMain () One { A     intN; -      while(cin>>n&&N) -     { theMemset (AN,0,sizeof(an)); -memset (ans,0,sizeof(ans)); -         intb; -          for(intI=1; i<=n;i++) +         { -scanf"%d%d",&a,&b); +an[a]+=1; Aan[b+1]+=-1; at         } -          for(intI=1; i<=n;i++) -         { -ans[i]=ans[i-1]+An[i]; -         } -cout<<ans[1]; in          for(intI=2; i<=n;i++) -         { tocout<<" "<<Ans[i]; +         } -cout<<Endl; the     } *     return 0; $}

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Simple method with a time complexity of n

Interval modification Point Query HDU1556