原题传送：http://acm.hdu.edu.cn/showproblem.php?pid=4417
划分树 + 二分。
总复杂度为mlog(n)log(n)，在O(10^7)左右，可以破了，二分查询区间第k大值，如果该值小于或等于给定高度，那么继续二分。那么最后得到的k值就是所求。
View Code
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 const int maxn = 100000 + 10; 5 6 int tree[20][maxn], num[20][maxn], a[maxn]; 7 8 void build(int l, int r, int d) 9 { 10 if(l == r) return ; 11 12 int mid = (l + r) >> 1, saved = 0, save = mid - l + 1; 13 int lp = l, rp = mid + 1; 14 for(int i = l; i <= r; i ++) 15 if(tree[d][i] < a[mid]) 16 --save; 17 for(int i = l; i <= r; i ++) 18 { 19 num[d][i] = (i == l ? 0 : num[d][i - 1]); 20 if(tree[d][i] < a[mid] || (saved < save && tree[d][i] == a[mid])) 21 { 22 ++ num[d][i]; 23 tree[d + 1][lp++] = tree[d][i]; 24 if(tree[d][i] == a[mid]) ++ saved; 25 } 26 else 27 tree[d + 1][rp++] = tree[d][i]; 28 } 29 build(l, mid, d + 1); 30 build(mid + 1, r, d + 1); 31 } 32 33 int query(int s, int t, int k, int l, int r, int d) 34 { 35 if(l == r) 36 return tree[d][s]; 37 38 int mid = (l + r) >> 1; 39 int s1 = num[d][s - 1]; 40 int s2 = num[d][t]; 41 if(s == l) // s == l时s-1到了另外的区间去了 42 s1 = 0; 43 44 if(s2 - s1 >= k) 45 return query(l + s1, l + s2 - 1, k, l, mid, d + 1); 46 47 int k1 = s - l - s1; 48 int k2 = (t - s + 1) - (s2 - s1); 49 return query(mid + 1 + k1, mid + k1 + k2, k - (s2 - s1), mid + 1, r, d + 1); 50 } 51 52 int solve(int s, int t, int h, int n) 53 { 54 int l = 1, r = (t - s) + 2, ans = 0; 55 while(l < r) 56 { 57 int mid = (l + r) >> 1; 58 int tmp = query(s, t, mid, 1, n, 0); 59 if(tmp <= h) 60 ans = mid, l = mid + 1; 61 else 62 r = mid; 63 } 64 return ans; 65 } 66 67 int main() 68 { 69 int T, n, m, u, v, w; 70 scanf("%d", &T); 71 for(int cas = 1; cas <= T; cas ++) 72 { 73 scanf("%d%d", &n, &m); 74 for(int i = 1; i <= n; i ++) 75 { 76 scanf("%d", &a[i]); 77 tree[0][i] = a[i]; 78 } 79 std::sort(a + 1, a + 1 + n); 80 build(1, n, 0); 81 printf("Case %d:\n", cas); 82 while(m --) 83 { 84 scanf("%d%d%d", &u, &v, &w); 85 printf("%d\n", solve(u + 1, v + 1, w, n)); 86 } 87 } 88 return 0; 89 }