这个题以前就做过了，用归并树+二分，2000+ms水过了。今天整理资料，zhk提到了划分树，于是去试了一下，经过几次小小的错误，最终过掉了，900+ms。时间效率很高，O(mlogn)，是区间查询第k大值的利器。
该总结的东西基本上都写在注释里面了，就不再废话了。
POJ 2104 K-th Number(划分树)
1 //从源序列开始，首先用buffer将此序列排序。
2 //每次划分的时候按照中点值将元素按照大小分别排在左子树和右子树，相当于一个排序的过程，用数组val记录。
3 //在划分的时候用数组toLeft记录每次划分时被分到左子树的元素的数目。
4
5 #include <stdio.h>
6 #include <string.h>
7 #include <algorithm>
8
9 const int MaxN = 100010;
10 int val[20][MaxN], toLeft[20][MaxN];
11 int srt[MaxN];
12 int M, N;
13
14 struct Node {
15 int l, r;
16 }node[MaxN<<3];
17
18 inline int L(int x) {return (x<<1)+1;}
19 inline int R(int x) {return (x<<1)+2;}
20
21 void build_tree(int s, int e, int cur_row, int id) {
22 node[id].l = s; node[id].r = e;
23 if(s == e)
24 return;
25 int mid = (s+e)>>1;
26 int emN = mid + 1 - s;
27 for(int i = s; i <= e; ++i) { //记录与中位数相同的元素被分到左子树的数目。
28 if(val[cur_row][i] < srt[mid]) //因为这些元素有可能被分到左子树，也有可能被分到右子树。
29 emN--; //记录这个值便于后面的操作。
30 }
31 int lp = s, rp = mid+1;
32 for(int i = s; i <= e; ++i) { //计算被划分到左子树的元素个数。
33 toLeft[cur_row][i] = i!=s?toLeft[cur_row][i-1]:0;
34 if(val[cur_row][i] < srt[mid]) {
35 val[cur_row+1][lp++] = val[cur_row][i];
36 toLeft[cur_row][i]++;
37 }
38 else if(val[cur_row][i] > srt[mid])
39 val[cur_row+1][rp++] = val[cur_row][i];
40 else if(emN) { //当emN计数为0时，划分到左子树的中位数元素划分完毕。
41 val[cur_row+1][lp++] = val[cur_row][i];
42 toLeft[cur_row][i]++;
43 emN--;
44 }
45 else
46 val[cur_row+1][rp++] = val[cur_row][i];
47 }
48 build_tree(s, mid, cur_row + 1, L(id)); //递归建树。
49 build_tree(mid+1, e, cur_row + 1, R(id));
50 }
51
52 //{[left part][query interval][right part]} 此为树的一个节点
53 //{[(LL)(RL)][(LI)(RI)][right part]}
54 // LL:划分到左子树的左边部分; RL:划分到右子树的左边部分
55 // LI:划分到左子树的中间部分; RL:划分到右子树的中间部分
56 //根据以上信息判断在哪个子树进行查询，并且重新计算query interval, 递归查询即可。
57
58 int query(int s, int e, int k, int cur_row, int id) {
59 if(s == e)
60 return val[cur_row][s];
61 int LL = s==node[id].l?0:toLeft[cur_row][s-1]; //计算LL, LI
62 int LI = toLeft[cur_row][e] - LL;
63 if(k <= LI)
64 return query(node[id].l+LL, node[id].l+LL+LI-1, k, cur_row+1, L(id));
65 int mid = (node[id].l + node[id].r) >> 1;
66 int RL = s - node[id].l - LL; //计算RL, RI
67 int RI = e + 1 - s - LI;
68 return query(mid+1+RL, mid+1+RL+RI-1, k - LI, cur_row+1, R(id));
69 }
70
71 int main() {
72 while(scanf("%d%d", &N, &M) == 2) {
73 for(int i = 0; i < N; ++i) {
74 scanf("%d", srt+i);
75 val[0][i] = srt[i];
76 }
77 std::sort(srt, srt + N);
78 build_tree(0, N-1, 0, 0);
79 int s, e, k;
80 for(int i = 0; i < M; ++i) {
81 scanf("%d%d%d", &s, &e, &k);
82 printf("%d\n", query(s-1, e-1, k, 0, 0));
83 }
84 }
85 return 0;
86 }
87
2 //每次划分的时候按照中点值将元素按照大小分别排在左子树和右子树，相当于一个排序的过程，用数组val记录。
3 //在划分的时候用数组toLeft记录每次划分时被分到左子树的元素的数目。
4
5 #include <stdio.h>
6 #include <string.h>
7 #include <algorithm>
8
9 const int MaxN = 100010;
10 int val[20][MaxN], toLeft[20][MaxN];
11 int srt[MaxN];
12 int M, N;
13
14 struct Node {
15 int l, r;
16 }node[MaxN<<3];
17
18 inline int L(int x) {return (x<<1)+1;}
19 inline int R(int x) {return (x<<1)+2;}
20
21 void build_tree(int s, int e, int cur_row, int id) {
22 node[id].l = s; node[id].r = e;
23 if(s == e)
24 return;
25 int mid = (s+e)>>1;
26 int emN = mid + 1 - s;
27 for(int i = s; i <= e; ++i) { //记录与中位数相同的元素被分到左子树的数目。
28 if(val[cur_row][i] < srt[mid]) //因为这些元素有可能被分到左子树，也有可能被分到右子树。
29 emN--; //记录这个值便于后面的操作。
30 }
31 int lp = s, rp = mid+1;
32 for(int i = s; i <= e; ++i) { //计算被划分到左子树的元素个数。
33 toLeft[cur_row][i] = i!=s?toLeft[cur_row][i-1]:0;
34 if(val[cur_row][i] < srt[mid]) {
35 val[cur_row+1][lp++] = val[cur_row][i];
36 toLeft[cur_row][i]++;
37 }
38 else if(val[cur_row][i] > srt[mid])
39 val[cur_row+1][rp++] = val[cur_row][i];
40 else if(emN) { //当emN计数为0时，划分到左子树的中位数元素划分完毕。
41 val[cur_row+1][lp++] = val[cur_row][i];
42 toLeft[cur_row][i]++;
43 emN--;
44 }
45 else
46 val[cur_row+1][rp++] = val[cur_row][i];
47 }
48 build_tree(s, mid, cur_row + 1, L(id)); //递归建树。
49 build_tree(mid+1, e, cur_row + 1, R(id));
50 }
51
52 //{[left part][query interval][right part]} 此为树的一个节点
53 //{[(LL)(RL)][(LI)(RI)][right part]}
54 // LL:划分到左子树的左边部分; RL:划分到右子树的左边部分
55 // LI:划分到左子树的中间部分; RL:划分到右子树的中间部分
56 //根据以上信息判断在哪个子树进行查询，并且重新计算query interval, 递归查询即可。
57
58 int query(int s, int e, int k, int cur_row, int id) {
59 if(s == e)
60 return val[cur_row][s];
61 int LL = s==node[id].l?0:toLeft[cur_row][s-1]; //计算LL, LI
62 int LI = toLeft[cur_row][e] - LL;
63 if(k <= LI)
64 return query(node[id].l+LL, node[id].l+LL+LI-1, k, cur_row+1, L(id));
65 int mid = (node[id].l + node[id].r) >> 1;
66 int RL = s - node[id].l - LL; //计算RL, RI
67 int RI = e + 1 - s - LI;
68 return query(mid+1+RL, mid+1+RL+RI-1, k - LI, cur_row+1, R(id));
69 }
70
71 int main() {
72 while(scanf("%d%d", &N, &M) == 2) {
73 for(int i = 0; i < N; ++i) {
74 scanf("%d", srt+i);
75 val[0][i] = srt[i];
76 }
77 std::sort(srt, srt + N);
78 build_tree(0, N-1, 0, 0);
79 int s, e, k;
80 for(int i = 0; i < M; ++i) {
81 scanf("%d%d%d", &s, &e, &k);
82 printf("%d\n", query(s-1, e-1, k, 0, 0));
83 }
84 }
85 return 0;
86 }
87
POJ 2104 K-th Number(归并+二分)
1 #include <stdio.h>
2 #include <string.h>
3 #include <algorithm>
4
5 using namespace std;
6
7 inline int MID(int s, int e) {return (s+e)>>1;}
8
9 const int MaxN = 100010;
10 int merge_tree[20][MaxN], n, m, dep[MaxN];
11 int u, v, k, flag;
12
13 void Merge(int begin, int end, int d) {
14 if(begin + 1 == end) {
15 merge_tree[d][begin] = merge_tree[0][begin];
16 dep[begin] = d;
17 return;
18 }
19 int Mid = MID(begin, end);
20 Merge(begin, Mid, d + 1);
21 Merge(Mid, end, d + 1);
22 int i, j, k;
23 for(i = begin, j = Mid, k = begin; i != Mid && j != end;) {
24 if(merge_tree[d+1][i] < merge_tree[d+1][j])
25 merge_tree[d][k++] = merge_tree[d+1][i++];
26 else
27 merge_tree[d][k++] = merge_tree[d+1][j++];
28 }
29 if(i != Mid) {
30 for(; i != Mid; ++i) merge_tree[d][k++] = merge_tree[d+1][i];
31 }
32 if(j != end) {
33 for(; j != end; ++j) merge_tree[d][k++] = merge_tree[d+1][j];
34 }
35 }
36
37 int search(int begin, int end, int s, int e, int x, int d) {
38 int Mid;
39 if(s == begin && end == e) {
40 if(x > merge_tree[d][end-1])
41 return end - begin;
42 if(x < merge_tree[d][begin])
43 return 0;
44 int u = begin - 1, v = end;
45 while(u+1 < v) {
46 Mid = MID(u, v);
47 if(merge_tree[d][Mid] <= x) {
48 if(merge_tree[d][Mid] == x)
49 flag++;
50 u = Mid;
51 }
52 else
53 v = Mid;
54 }
55 return u - begin + 1;
56 }
57 Mid = MID(begin, end);
58 if(s >= Mid)
59 return search(Mid, end, s, e, x, d+1);
60 if(e <= Mid)
61 return search(begin, Mid, s, e, x, d+1);
62 return search(begin, Mid, s, Mid, x, d+1) + search(Mid, end, Mid, e, x, d+1);
63 }
64
65 void work() {
66 int begin = -1, end = n, Mid;
67 while(begin + 1 < end) {
68 Mid = MID(begin, end);
69 flag = 0;
70 int sum = search(0, n, u, v+1, merge_tree[0][Mid], 0);
71 if(flag && sum - flag < k && sum >= k) {
72 printf("%d\n", merge_tree[0][Mid]);
73 return;
74 }
75 else if(sum < k)
76 begin = Mid;
77 else
78 end = Mid;
79 }
80 }
81
82 int main() {
83 while(scanf("%d%d", &n, &m) != EOF) {
84 for(int i = 0; i < n; ++i) scanf("%d", merge_tree[0] + i);
85 Merge(0, n, 0);
86 for(int i = 0; i < m; ++i) {
87 scanf("%d%d%d", &u, &v, &k);
88 u--; v--;
89 work();
90 }
91 }
92 return 0;
93 }
94
2 #include <string.h>
3 #include <algorithm>
4
5 using namespace std;
6
7 inline int MID(int s, int e) {return (s+e)>>1;}
8
9 const int MaxN = 100010;
10 int merge_tree[20][MaxN], n, m, dep[MaxN];
11 int u, v, k, flag;
12
13 void Merge(int begin, int end, int d) {
14 if(begin + 1 == end) {
15 merge_tree[d][begin] = merge_tree[0][begin];
16 dep[begin] = d;
17 return;
18 }
19 int Mid = MID(begin, end);
20 Merge(begin, Mid, d + 1);
21 Merge(Mid, end, d + 1);
22 int i, j, k;
23 for(i = begin, j = Mid, k = begin; i != Mid && j != end;) {
24 if(merge_tree[d+1][i] < merge_tree[d+1][j])
25 merge_tree[d][k++] = merge_tree[d+1][i++];
26 else
27 merge_tree[d][k++] = merge_tree[d+1][j++];
28 }
29 if(i != Mid) {
30 for(; i != Mid; ++i) merge_tree[d][k++] = merge_tree[d+1][i];
31 }
32 if(j != end) {
33 for(; j != end; ++j) merge_tree[d][k++] = merge_tree[d+1][j];
34 }
35 }
36
37 int search(int begin, int end, int s, int e, int x, int d) {
38 int Mid;
39 if(s == begin && end == e) {
40 if(x > merge_tree[d][end-1])
41 return end - begin;
42 if(x < merge_tree[d][begin])
43 return 0;
44 int u = begin - 1, v = end;
45 while(u+1 < v) {
46 Mid = MID(u, v);
47 if(merge_tree[d][Mid] <= x) {
48 if(merge_tree[d][Mid] == x)
49 flag++;
50 u = Mid;
51 }
52 else
53 v = Mid;
54 }
55 return u - begin + 1;
56 }
57 Mid = MID(begin, end);
58 if(s >= Mid)
59 return search(Mid, end, s, e, x, d+1);
60 if(e <= Mid)
61 return search(begin, Mid, s, e, x, d+1);
62 return search(begin, Mid, s, Mid, x, d+1) + search(Mid, end, Mid, e, x, d+1);
63 }
64
65 void work() {
66 int begin = -1, end = n, Mid;
67 while(begin + 1 < end) {
68 Mid = MID(begin, end);
69 flag = 0;
70 int sum = search(0, n, u, v+1, merge_tree[0][Mid], 0);
71 if(flag && sum - flag < k && sum >= k) {
72 printf("%d\n", merge_tree[0][Mid]);
73 return;
74 }
75 else if(sum < k)
76 begin = Mid;
77 else
78 end = Mid;
79 }
80 }
81
82 int main() {
83 while(scanf("%d%d", &n, &m) != EOF) {
84 for(int i = 0; i < n; ++i) scanf("%d", merge_tree[0] + i);
85 Merge(0, n, 0);
86 for(int i = 0; i < m; ++i) {
87 scanf("%d%d%d", &u, &v, &k);
88 u--; v--;
89 work();
90 }
91 }
92 return 0;
93 }
94
