划分树 hdu4417Super Mario

划分树+二分枚举
http://acm.hdu.edu.cn/showproblem.php?pid=4417
划分树http://www.cnblogs.com/pony1993/archive/2012/07/17/2594544.html
直接搬来模板打得
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  1 #include <iostream>
  2  #include <cstdio>
  3  #include <algorithm>
  4  using namespace std;
  5  #define N 100005
  6  int a[N], as[N];//原数组，排序后数组
  7  int n, m;
  8  int sum[20][N];//记录第i层的1~j划分到左子树的元素个数(包括j)
  9  int tree[20][N];//记录第i层元素序列
 10  void build(int c, int l, int r){
 11      int i, mid = (l + r) >> 1, lm = mid - l + 1, lp = l, rp = mid + 1;
 12      for (i = l; i <= mid; i++){
 13          if (as[i] < as[mid]){
 14              lm--;//先假设左边的(mid - l + 1)个数都等于as[mid],然后把实际上小于as[mid]的减去
 15          }
 16      }
 17      for (i = l; i <= r; i++){
 18          if (i == l){
 19              sum[c][i] = 0;//sum[i]表示[l, i]内有多少个数分到左边，用DP来维护
 20          }else{
 21              sum[c][i] = sum[c][i - 1];
 22          }
 23          if (tree[c][i] == as[mid]){
 24              if (lm){
 25                  lm--;
 26                  sum[c][i]++;
 27                  tree[c + 1][lp++] = tree[c][i];
 28              }else
 29                  tree[c + 1][rp++] = tree[c][i];
 30          } else if (tree[c][i] < as[mid]){
 31              sum[c][i]++;
 32              tree[c + 1][lp++] = tree[c][i];
 33          } else{
 34              tree[c + 1][rp++] = tree[c][i];
 35          }
 36      }
 37      if (l == r)return;
 38      build(c + 1, l, mid);
 39      build(c + 1, mid + 1, r);
 40  }
 41  int query(int c, int l, int r, int ql, int qr, int k){
 42      int s;//[l, ql)内将被划分到左子树的元素数目
 43      int ss;//[ql, qr]内将被划分到左子树的元素数目
 44      int mid = (l + r) >> 1;
 45      if (l == r){
 46          return tree[c][l];
 47      }
 48      if (l == ql){//这里要特殊处理！
 49      s = 0;
 50      ss = sum[c][qr];
 51      }else{
 52          s = sum[c][ql - 1];
 53          ss = sum[c][qr] - s;
 54      }//假设要在区间[l,r]中查找第k大元素，t为当前节点，lch，rch为左右孩子，left，mid为节点t左边界和中间点。
 55      if (k <= ss){//sum[r]-sum[l-1]>=k，查找lch[t],区间对应为[ left+sum[l-1], left+sum[r]-1 ]
 56          return query(c + 1, l, mid, l + s, l + s + ss - 1, k);
 57      }else{//sum[r]-sum[l-1]<k,查找rch[t]，区间对应为[ mid+1+l-left-sum[l-1], mid+1+r-left-sum[r] ]
 58          return query(c + 1, mid + 1, r, mid - l + 1 + ql - s, mid - l + 1 + qr - s - ss,k - ss);
 59      }
 60  }
 61  int main()
 62  {
 63      int i, j, k,t,kk=0;
 64      scanf("%d",&t);
 65      while(t--)
 66      {
 67          kk++;
 68          scanf("%d%d", &n, &m);
 69          for (i = 1; i <= n; i++)
 70          {
 71              scanf("%d", &a[i]);
 72              tree[0][i] = as[i] = a[i];
 73          }
 74          sort(as + 1, as + 1 + n);
 75          build(0, 1, n);
 76          printf("Case %d:\n",kk);
 77          while(m--)
 78          {
 79              scanf("%d%d%d",&i,&j,&k);
 80              i++;
 81              j++;
 82              int low = 1;
 83              int high = j-i+1;
 84              int re = 0;
 85              while(low<=high)
 86              {
 87                  int m = (low+high)>>1;
 88                  int w = query(0, 1, n, i, j, m);
 89                  if(w<=k)
 90                  {
 91                      low = m+1;
 92                      re = m;
 93                  }
 94                  else
 95                  high = m-1;
 96              }
 97              printf("%d\n",re);
 98          }
 99      }
100      return 0;
101  }