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HDU 2665.Kth number 区间第K小
Kth number
Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11394 Accepted Submission(s): 3465
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
HDU男生专场公开赛——赶在女生之前先过节（From WHU）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2665
题意：求区间第K小（区间升序第k个数）
思路：归并树。二分答案x。判断不超过x的数在区间内多少个。假设x是区间第k小，那么在区间内不超过x的数不少于k个。
代码：
区间第K小
#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<stack>#include<vector>#include<set>using namespace std;#define PI acos(-1.0)typedef long long ll;typedef pair<int,int> P;const int maxn=1e5+100,maxm=1e5+100,inf=0x3f3f3f3f,mod=1e9+7;const ll INF=1e13+7;inline int get_int(){ int num=0; char ch; while((ch=getchar())!=‘ ‘&&ch!=‘\n‘) num=num*10+(ch-‘0‘); return num;}/****************************/struct edge{ int from,to; int cost;};edge es[maxm];struct node{ int num; int k;};node sign[maxn];int a[maxn];int cmp(node x,node y){ return x.num<y.num;}vector<node>tree[maxn<<2];void build(int l,int r,int pos){ if(l==r) return; int mid=(l+r)/2; for(int i=0; i<tree[pos].size(); i++) { int k=tree[pos][i].k; if(l<=k&&k<=mid) tree[pos<<1].push_back(tree[pos][i]); else tree[pos<<1|1].push_back(tree[pos][i]); } build(l,mid,pos<<1); build(mid+1,r,pos<<1|1);}int query(int L,int R,int w,int l,int r,int pos){ if(L<=l&&r<=R) { int s=0,e=tree[pos].size()-1; int cou=-1; while(s<=e) { int md=(s+e)/2; if(tree[pos][md].num<=w) s=md+1,cou=md; else e=md-1; } return cou+1; } int mid=(l+r)/2; int ans=0; if(L<=mid) ans+=query(L,R,w,l,mid,pos<<1); if(R>mid) ans+=query(L,R,w,mid+1,r,pos<<1|1); return ans;}int main(){ int T; scanf("%d",&T); while(T--) { int n,q; scanf("%d%d",&n,&q); for(int i=0; i<=4*n; i++) tree[i].clear(); for(int i=0; i<n; i++) { scanf("%d",&a[i]); sign[i].num=a[i],sign[i].k=i+1; } sort(sign,sign+n,cmp); for(int i=0; i<n; i++) tree[1].push_back(sign[i]); build(1,n,1); while(q--) { int l,r,k; scanf("%d%d%d",&l,&r,&k); int L=0,R=n-1; int ans=n-1; while(L<=R) { int mid=(L+R)/2; int w=sign[mid].num; if((query(l,r,w,1,n,1))>=k) R=mid-1,ans=mid; else L=mid+1; } printf("%d\n",sign[ans].num); } } return 0;}
HDU 2665.Kth number 区间第K小
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