HDU 5412 CRB and Queries 动态整体二分

Problem Description
There are N boys in CodeLand.
Boy i has his coding skill Ai.
CRB wants to know who has the suitable coding skill.
So you should treat the following two types of queries.
Query 1: 1 l v
The coding skill of Boy l has changed to v.
Query 2: 2 l r k
This is a report query which asks the k-th smallest value of coding skill between Boy l and Boy r(both inclusive).
Input
There are multiple test cases.
The first line contains a single integer N.
Next line contains N space separated integers A1, A2, …, AN, where Ai denotes initial coding skill of Boy i.
Next line contains a single integer Q representing the number of queries.
Next Q lines contain queries which can be any of the two types.
1 ≤ N, Q ≤ 105
1 ≤ Ai, v ≤ 109
1 ≤ l ≤ r ≤ N
1 ≤ k ≤ r – l + 1
 
Output
For each query of type 2, output a single integer corresponding to the answer in a single line.
 
Sample Input
5 1 2 3 4 5 3 2 2 4 2 1 3 6 2 2 4 2
Sample Output
3 4
 
题解:
按时间顺序加入,所有的修改拆成删除和添加两个操作,分别对应树状数组中的加减.
维护时间顺序即可
  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #include <cstdio>
  6 #include <cmath>
  7 using namespace std;
  8 const int N=100005;
  9 int gi(){
 10     int str=0;char ch=getchar();
 11     while(ch>'9' || ch<'0')ch=getchar();
 12     while(ch>='0' && ch<='9')str=(str<<1)+(str<<3)+ch-48,ch=getchar();
 13     return str;
 14 }
 15 int n,m,a[N],top=0,ans[N];
 16 struct node{
 17     int x,y,cnt,k,id;
 18 }t[N*4],q1[N*4],q2[N*4];
 19 int Tree[N*4];
 20 void add(int sta,int ad){
 21     for(int i=sta;i<=n;i+=(i&(-i)))Tree[i]+=ad;
 22 }
 23 int getsum(int sta){
 24     int sum=0;
 25     for(int i=sta;i>=1;i-=(i&(-i)))sum+=Tree[i];
 26     return sum;
 27 }
 28 int l1=0,l2=0;
 29 void count(int ll,int rr,int dl,int dr)
 30     {
 31         for(int i=ll;i<=rr;i++)
 32             {
 33                 if(t[i].k)
 34                     t[i].cnt=getsum(t[i].y)-getsum(t[i].x-1);
 35                 else if(t[i].y<=dr)
 36                     add(t[i].x,t[i].cnt);
 37             }
 38         for(int i=ll;i<=rr;i++)
 39             if(!t[i].k && t[i].y<=dr)
 40                 add(t[i].x,-t[i].cnt);
 41         l1=0;l2=0;
 42         for(int i=ll;i<=rr;i++)
 43             {
 44                 if(t[i].k)
 45                     {
 46                         if(t[i].cnt>=t[i].k)
 47                             q1[++l1]=t[i];
 48                         else
 49                             t[i].k-=t[i].cnt,q2[++l2]=t[i];
 50                     }
 51                 else
 52                     {
 53                         if(t[i].y<=dr)
 54                             q1[++l1]=t[i];
 55                         else
 56                             q2[++l2]=t[i];
 57                     }
 58             }
 59         int now=ll-1;
 60         for(int i=1;i<=l1;i++)
 61             t[++now]=q1[i];
 62         for(int i=1;i<=l2;i++)
 63             t[++now]=q2[i];
 64    }
 65 void div(int ll,int rr,int dl,int dr)
 66     {
 67         if(dl==dr)
 68             {
 69                 for(int i=ll;i<=rr;i++)
 70                     if(t[i].k)ans[t[i].id]=dl;
 71                 return ;
 72             }
 73         int mid=(dl+dr)>>1;
 74         count(ll,rr,dl,mid);
 75         int to=l1;
 76         if(to)
 77             div(ll,ll+to-1,dl,mid);
 78         if(to<=rr-ll)
 79             div(ll+to,rr,mid+1,dr);
 80     }
 81 void Clear(){
 82     top=0;
 83     memset(ans,0,sizeof(ans));
 84     memset(t,0,sizeof(t));
 85 }
 86 void work()
 87     {
 88         Clear();
 89        for(int i=1;i<=n;i++)
 90         {
 91             a[i]=gi();
 92             t[++top]=((node){i,a[i],1,0,0});
 93         }
 94         m=gi();
 95        int flag,x,y,z;
 96       for(int i=1;i<=m;i++)
 97         {
 98             flag=gi();
 99             if(flag==1)
100                 {
101                     x=gi();y=gi();
102                     t[++top]=((node){x,a[x],-1,0,0});
103                     t[++top]=((node){x,y,1,0,0});
104                     a[x]=y;
105                 }
106             else
107                 {
108                     x=gi();y=gi();z=gi();
109                     t[++top]=(node){x,y,0,z,i};
110                 }
111         }
112        div(1,top,1,1e9);
113        for(int i=1;i<=m;i++)if(ans[i])printf("%d\n",ans[i]);
114     }
115 int main()
116 {
117     while(~scanf("%d",&n))
118         work();
119     return 0;
120 }

 

还有主席树维护的树状数组，但是MLE,这种题目如果数据小可以用

 

  1 #include <algorithm>
  2 #include <iostream>
  3 #include <cstdlib>
  4 #include <cstring>
  5 #include <cstdio>
  6 #include <cmath>
  7 using namespace std;
  8 const int N=100001;
  9 int n,m,a[N],b[N<<1];
 10 struct Ques{
 11     int flag,x,y,k;
 12 }q[N];
 13 int num=0,bel[N];
 14 int getrank(int x){
 15     int l=1,r=num,mid;
 16     while(l<=r)
 17         {
 18             mid=(l+r)>>1;
 19             if(b[mid]==x)return mid;
 20             if(x>b[mid])l=mid+1;
 21             else r=mid-1;
 22         }
 23 }
 24 struct Tree{
 25     int ls,rs,cnt;
 26 }t[N*600];
 27 int root[N],tot=0;
 28 void add(int &rt,int l,int r,int val,int ad)
 29     {
 30         t[++tot]=t[rt];rt=tot;t[rt].cnt+=ad;
 31         if(l==r)return ;
 32         int mid=(l+r)>>1;
 33         if(val<=mid)add(t[rt].ls,l,mid,val,ad);
 34         else add(t[rt].rs,mid+1,r,val,ad);
 35     }
 36 void updata(int sta,int val,int ad)
 37     {
 38         for(int i=sta;i<=n;i+=(i&(-i)))add(root[i],1,num,val,ad);
 39     }
 40 int nl[N],nr[N];
 41 int query(int ll,int rr,int rank)
 42     {
 43         int l=1,r=num,mid;
 44         int lenl=0,lenr=0,sl,sr;
 45         for(int i=ll;i>=1;i-=(i&(-i)))nl[++lenl]=root[i];
 46         for(int i=rr;i>=1;i-=(i&(-i)))nr[++lenr]=root[i];
 47         while(l<r)
 48             {
 49                 mid=(l+r)>>1;
 50                sl=sr=0;
 51                 for(int i=1;i<=lenl;i++)sl+=t[t[nl[i]].ls].cnt;
 52                 for(int i=1;i<=lenr;i++)sr+=t[t[nr[i]].ls].cnt;
 53                 if(sr-sl>=rank)
 54                     {
 55                         for(int i=1;i<=lenl;i++)nl[i]=t[nl[i]].ls;
 56                         for(int i=1;i<=lenr;i++)nr[i]=t[nr[i]].ls;
 57                         r=mid;
 58                     }
 59                 else
 60                     {
 61                         for(int i=1;i<=lenl;i++)nl[i]=t[nl[i]].rs;
 62                         for(int i=1;i<=lenr;i++)nr[i]=t[nr[i]].rs;
 63                         l=mid+1;rank-=sr-sl;
 64                     }
 65             }
 66         return b[r];
 67     }
 68 void Clear()
 69     {
 70         tot=0;num=0;
 71         memset(root,0,sizeof(root));
 72         memset(t,0,sizeof(t));
 73     }
 74 void work()
 75     {
 76         Clear();
 77         for(int i=1;i<=n;i++)scanf("%d",&a[i]),b[++num]=a[i];
 78         scanf("%d",&m);
 79         for(int i=1;i<=m;i++)
 80             {
 81                 scanf("%d",&q[i].flag);
 82                 if(q[i].flag==2){
 83                     scanf("%d%d%d",&q[i].x,&q[i].y,&q[i].k);
 84                 }
 85                 else{
 86                     scanf("%d%d",&q[i].x,&q[i].y);
 87                     b[++num]=q[i].y;
 88                 }
 89             }
 90         sort(b+1,b+num+1);
 91         for(int i=1;i<=n;i++)bel[i]=getrank(a[i]);
 92         for(int i=1;i<=n;i++)updata(i,bel[i],1);
 93         for(int i=1;i<=m;i++)
 94             {
 95                 if(q[i].flag==1)
 96                     {
 97                         updata(q[i].x,bel[q[i].x],-1);
 98                         updata(q[i].x,q[i].y=getrank(q[i].y),1);
 99                         bel[q[i].x]=q[i].y;
100                     }
101                     else
102                      printf("%d\n",query(q[i].x-1,q[i].y,q[i].k));
103             }
104     }
105 int main()
106 {
107     //freopen("pp.in","r",stdin);
108     //freopen("pp.out","w",stdout);
109     while(~scanf("%d",&n))
110         work();
111     return 0;
112 }

 

 

 

posted @ 2017-07-03 15:31  PIPIBoss  阅读(282)  评论(0)  编辑  收藏  举报