Kth number
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10682 Accepted Submission(s): 3268
Problem Description
Give you a sequence and ask you the kth big number of a inteval.
Input
The first line is the number of the test cases.
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]
For each test case, the first line contain two integer n and m (n, m <= 100000), indicates the number of integers in the sequence and the number of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]
Output
For each test case, output m lines. Each line contains the kth big number.
Sample Input
1
10 1
1 4 2 3 5 6 7 8 9 0
1 3 2
Sample Output
2
Source
说实话 我不知道我该怎么去解释 因为这就是主席树的板子
无区间修改的主席树 不解释
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cctype> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<stack> 9 #include<set> 10 #include<vector> 11 #include<algorithm> 12 #include<string.h> 13 #define ll long long 14 #define LL unsigned long long 15 using namespace std; 16 const int INF=0x3f3f3f3f; 17 const double eps=0.0000000001; 18 const int N=100000+10; 19 struct node{ 20 int left,right; 21 int val; 22 }tree[N*40]; 23 int cnt; 24 int a[N],b[N]; 25 int root[N]; 26 int build(int l,int r){ 27 int pos=++cnt; 28 tree[pos].val=0; 29 if(l==r)return pos; 30 int mid=(l+r)>>1; 31 tree[pos].left=build(l,mid); 32 tree[pos].right=build(mid+1,r); 33 return pos; 34 } 35 void update(int pre,int &now,int x,int l,int r){ 36 tree[++cnt]=tree[pre]; 37 now=cnt; 38 tree[now].val++; 39 if(l==r)return; 40 int mid=(l+r)>>1; 41 if(x<=mid){ 42 update(tree[pre].left,tree[now].left,x,l,mid); 43 } 44 else{ 45 update(tree[pre].right,tree[now].right,x,mid+1,r); 46 } 47 48 } 49 int query(int L,int R,int k,int l,int r){ 50 if(l==r) return l; 51 int mid=(l+r)>>1; 52 int sum=tree[tree[R].left].val-tree[tree[L].left].val; 53 if(k<=sum){ 54 return query(tree[L].left,tree[R].left,k,l,mid); 55 } 56 else{ 57 return query(tree[L].right,tree[R].right,k-sum,mid+1,r); 58 } 59 60 } 61 int main() 62 { 63 int t; 64 scanf("%d",&t); 65 while(t--){ 66 memset(root,0,sizeof(root)); 67 memset(a,0,sizeof(a)); 68 memset(b,0,sizeof(b)); 69 int n,m; 70 scanf("%d%d",&n,&m); 71 cnt=0; 72 for(int i=1;i<=n;i++) { 73 scanf("%d",&a[i]); 74 b[i]=a[i]; 75 } 76 sort(b+1,b+1+n); 77 int tt = unique(b+1,b+1+n)-b-1;
root[0]=build(1,tt); 78 for(int i=1;i<=n;i++) { 79 a[i]=lower_bound(b+1,b+1+tt,a[i])-b;//二分找到a[i]的位置 81 update(root[i-1],root[i],a[i],1,tt);//root[i-1]表示上一个版本的线段树 82 } 83 for(int i=1;i<=m;i++) { 84 int l,r,k; 85 scanf("%d%d%d",&l,&r,&k); 86 int ans=query(root[l-1],root[r],k,1,tt); 87 printf("%d ", b[ans]); 88 } 89 } 90 return 0; 91 }