【MD-80】【hdu】Kth number

Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12220 Accepted Submission(s): 3709

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]

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）

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题解

在主席生日后不久，我们就收到了主席树的作业

google翻译：给你一个序列，并问你第k个大数字的一个。
题意：从小到大排序后第k个数
SMG！！！
然后主席树对于每一个操作都会新开一棵值域线段树树，并把没有改变的结点连到前一棵树上去
这样算来空间复杂度O（n*logINF）
轰！
所以要离散化（喵喵喵）
于是变为O（nlogn）
ps：把多组数据去掉可以交poj2104

ac代码

#include<cstdio>#include<iostream>#include<algorithm>#define N 100010#define INF 2147483647using namespace std;int t[N*20],num,lc[N*20],rc[N*20],n,m,p,q,o,T,rt[N],b[N];struct Node{    int v,id;}a[N];bool cmp(Node p,Node q){    return p.v<q.v;}void Insert(int& x,int y,int l,int r){    x=++num;    t[x]=t[y]+1;    lc[x]=lc[y];    rc[x]=rc[y];    if(l==r)return ;    int mid=(l+r)>>1;    if(p<=mid)Insert(lc[x],lc[y],l,mid);    else Insert(rc[x],rc[y],mid+1,r);}int Query(int x,int y,int l,int r,int k){    if(l==r)return a[l].v;    int mid=(l+r)>>1;    int d=t[lc[x]]-t[lc[y]];    if(k<=d)return Query(lc[x],lc[y],l,mid,k);    else return Query(rc[x],rc[y],mid+1,r,k-d);}int main(){    freopen("data.txt","r",stdin);    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        num=0;        for(int i=1;i<=n;i++){            scanf("%d",&a[i].v);            a[i].id=i;        }        sort(a+1,a+n+1,cmp);        for(int i=1;i<=n;i++)b[a[i].id]=i;        for(int i=1;i<=n;i++){            p=b[i];            Insert(rt[i],rt[i-1],1,n);        }        for(int i=1;i<=m;i++){            scanf("%d%d%d",&p,&q,&o);            printf("%d\n",Query(rt[q],rt[p-1],1,n,o));        }    }}

对拍

#include<cstdio>#include<iostream>#include<ctime>#include<cstdlib>#include<algorithm>#define MOD 100010using namespace std;int n,a[MOD+10],m,p,q,T;int main(){    freopen("data.txt","w",stdout);    srand((unsigned)time(NULL));    T=n=rand()%MOD+1;    printf("%d\n",T);    while(T--){        n=rand()%MOD+1;        m=rand()%MOD+1;        printf("%d %d\n",n,m);        for(int i=1;i<=n;i++)a[i]=rand()%MOD+1;        for(int i=1;i<=n;i++)printf("%d ",a[i]);        printf("\n");        for(int i=1;i<=m;i++){            while((p=rand()%n+1)>(q=rand()%n+1));            printf("%d %d %d\n",p,q,rand()%(q-p+1)+1);        }    }}

暴力

sort大暴力

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#define N 200010using namespace std;int a[N],n,m,p,q,o,b[N],T;bool cmp(int p,int q){    return p>q;}int main(){    freopen("data.txt","r",stdin);    freopen("2.txt","w",stdout);    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        memcpy(b,a,sizeof(a));        for(int i=1;i<=m;i++){            memcpy(b,a,sizeof(a));            scanf("%d%d%d",&p,&q,&o);            sort(b+p,b+q+1,cmp);            printf("%d\n",b[p+o-1]);        }    }}