1087 - Diablo

 

1087 - Diablo

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Time Limit: 2 second(s)

Memory Limit: 64 MB

All of you must have played the game 'Diablo'. It's an exclusive game to play. In this game the main opponent of you is Diablo. If you kill him the game finishes. But as usual, Diablo is smarter than you.

Diablo has a large number of army. Diablo arranges them in a line in any arbitrary order and everyone is given an integer id. Each time Diablo either adds one army in the end or he calls for the k^th army (from left) from the line. Then the army gets out and it attacks you.

Since you are a great magician, you can read Diablo's mind. Now you want to find the id of the armies who are about to attack you.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

The first line of each case is a blank line. The next line contains two integers n (0 ≤ n ≤ 10⁵), denoting the number of the initial army and q (1 ≤ q ≤ 50000) representing the number of queries. The next line contains n space separated integers. The i^th integer of this line denotes the id of the i^th person. Each of these integers will be positive and fits into a 32 bit signed integer. Each of the next q lines will contain a query, of the form:

a p    (add a person at the end of the line whose id is p)

c k    (call the k^th person from the line (from left), k is a positive 32 bit signed integer)

Output

For each case of input, print the case number in a line. Then for all the queries 'c k' you have to print the id of the k^th person or 'none' if there is none.

Sample Input	Output for Sample Input
2 5 5 6 5 3 2 1 c 1 c 1 a 20 c 4 c 4 2 1 18811 1991 c 1	Case 1: 6 5 20 none Case 2: 18811

Notes

Dataset is huge, use faster i/o methods.

---

PROBLEM SETTER: JANE ALAM JAN

思路：线段数求第k大树，单点更新；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<stack>
  7 #include<map>
  8 #include<math.h>
  9 using namespace std;
 10 typedef long long LL;
 11 int tree[5000000];
 12 int ans[160005];
 13 int id[160000];
 14 int flag[160000];
 15 void build(int l,int r,int k,int u);
 16 int ask(int l,int r,int k,int m);
 17 void up(int k);
 18 void in(int x);
 19 int main(void)
 20 {
 21         int i,j,k;
 22         scanf("%d",&k);
 23         int s;
 24         int n,m;
 25         int gg=0;
 26         for(s=1; s<=k; s++)
 27         {
 28                 scanf("%d %d",&n,&m);
 29                 for(i=1; i<=n; i++)
 30                 {
 31                         scanf("%d",&ans[i]);
 32                 }
 33                 memset(flag,0,sizeof(flag));
 34                 for(i=n+1; i<=160000; i++)
 35                 {
 36                         flag[i]=1;
 37                 }
 38                 build(1,160000,0,n);
 39                 gg=n;
 40                 int cn=n;
 41                 printf("Case %d:\n",s);
 42                 while(m--)
 43                 {
 44                         char a[2];
 45                         int x;
 46                         scanf("%s %d",a,&x);
 47                         if(a[0]=='c')
 48                         {
 49                                 if(gg<x)
 50                                         printf("none\n");
 51                                 else
 52                                 {
 53                                         int cc=ask(1,160000,0,x);
 54                                         printf("%d\n",ans[cc]);
 55                                         gg--;
 56                                 }
 57                         }
 58                         else
 59                         {
 60                                 cn++;
 61                                 flag[cn]=0;
 62                                 ans[cn]=x;
 63                                 gg++;
 64                                 in(cn);
 65                         }
 66                 }
 67         }
 68         return 0;
 69 }
 70 void build(int l,int r,int k,int u)
 71 {
 72         if(l==r)
 73         {
 74                 if(r<=u)
 75                 {
 76                         id[l]=k;
 77                         tree[k]=1;
 78                 }
 79                 else
 80                 {
 81                         id[l]=k;
 82                         tree[k]=0;
 83                 }
 84                 return ;
 85         }
 86         else
 87         {
 88                 build(l,(l+r)/2,2*k+1,u);
 89                 build((l+r)/2+1,r,2*k+2,u);
 90                 tree[k]=tree[2*k+1]+tree[2*k+2];
 91         }
 92 }
 93 int ask(int l,int r,int k,int ans)
 94 {
 95         if(ans==tree[k])
 96         {
 97                 int c=id[r];
 98                 if(flag[r]==0)
 99                 {  flag[r]=1;
100                         up(c);
101                         return r;
102                 }
103                 else
104                 {
105                         if(tree[2*k+1]<ans)
106                         {
107                                 return ask((l+r)/2+1,r,2*k+2,ans-tree[2*k+1]);
108                         }
109                         else if(tree[2*k+1]==ans)
110                         {
111                                 return ask(l,(l+r)/2,2*k+1,ans);
112                         }
113                 }
114         }
115         else  if(ans<tree[k])
116         {
117                 if(tree[2*k+1]>=ans)
118                 {
119                         return ask(l,(l+r)/2,2*k+1,ans);
120                 }
121                 else if(tree[2*k+1]<ans)
122                 {
123                         return ask((l+r)/2+1,r,2*k+2,ans-tree[2*k+1]);
124                 }
125         }
126 }
127 void up(int k)
128 {
129         tree[k]=0;
130         if(k==0)return ;
131         k=(k-1)/2;
132         while(k>=0)
133         {
134                 tree[k]=tree[2*k+1]+tree[2*k+2];
135                 if(k==0)
136                         return ;
137                 k=(k-1)/2;
138         }
139 }
140 void in(int x)
141 {
142         int cc=id[x];
143         tree[cc]=1;
144         if(cc==0)return ;
145         cc=(cc-1)/2;
146         while(cc>=0)
147         {
148                 tree[cc]=tree[2*cc+1]+tree[2*cc+2];
149                 if(cc==0)
150                         return ;
151                 cc=(cc-1)/2;
152         }
153 }