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  }