poj2761 Feed the dogs【解法一】

Description Wind loves pretty dogs very much, and she has n pet dogs.
So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind,
but not the dogs, so Jiajia use a special way to feed the dogs. At
lunchtime, the dogs will stand on one line, numbered from 1 to n, the
leftmost one is 1, the second one is 2, and so on. In each feeding,
Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of
course Jiajia has his own way of deciding the pretty value of each
dog. It should be noted that Jiajia do not want to feed any position
too much, because it may cause some death of dogs. If so, Wind will be
angry and the aftereffect will be serious. Hence any feeding inteval
will not contain another completely, though the intervals may
intersect with each other.

Your task is to help Jiajia calculate which dog ate the food after
each feeding.

Input The first line contains n and m, indicates the number of dogs
and the number of feedings.

The second line contains n integers, describe the pretty value of each
dog from left to right. You should notice that the dog with lower
pretty value is prettier.

Each of following m lines contain three integer i,j,k, it means that
Jiajia feed the k-th pretty dog in this feeding.

You can assume that n<100001 and m<50001.

Output Output file has m lines. The i-th line should contain the
pretty value of the dog who got the food in the i-th feeding.

解法二见【这里】。
离散化以后按照数字大小维护树状数组，把询问按区间从左到右排序之后线性扫描，维护当前区间内数字个数，再二分答案查找第k小即可。总时间复杂度O(nlogn+mlognlogn)。

#include
#include
#include
using namespace std;
struct query
{
    int l,r,k,num;
    bool operator < (const query & qqq) const
    {
        return lq[50010];
int ori[100010],a[100010],ord[100010],m,n,nn,s[100010],ans[50010];
void inc(int x)
{
    for (;x<=nn;x+=x&-x)
      s[x]++;
}
void dec(int x)
{
    for (;x<=nn;x+=x&-x)
      s[x]--;
}
int sum(int x)
{
    int ret=0;
    for (;x;x-=x&-x)
      ret+=s[x];
    return ret;
}
int find(int k)
{
    int l,r,mid,i,j,x;
    l=1;
    r=nn;
    while (l2;
        if (sum(mid)>=k) r=mid;
        else l=mid+1;
    }
    return l;
}
int main()
{
    int i,j,k,p,x,y,z;
    scanf("%d%d",&n,&m);
    for (i=1;i<=n;i++)
    {
        scanf("%d",&ori[i]);
        ord[i]=ori[i];
    }
    sort(ord+1,ord+n+1);
    nn=unique(ord+1,ord+n+1)-ord-1;
    for (i=1;i<=n;i++)
      a[i]=lower_bound(ord+1,ord+nn+1,ori[i])-ord;
    for (i=1;i<=m;i++)
    {
        scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].k);
        q[i].num=i;
    }
    sort(q+1,q+m+1);
    q[0].l=q[1].l;
    q[0].r=q[1].l-1;
    for (i=1;i<=m;i++)
    {
        for (j=q[i-1].l;j<q[i].l;j++)
          dec(a[j]);
        for (j=q[i-1].r+1;j<=q[i].r;j++)
          inc(a[j]);
        ans[q[i].num]=find(q[i].k);
    }
    for (i=1;i<=m;i++)
      printf("%d\n",ord[ans[i]]);
}