zoj 2112 Dynamic Rankings（k大数水过）

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1112

　　用第k大数8s水过的题....如果计算复杂度，理论上是不应该可以通过的...这题正解是树套树，用线段树嵌套平衡树。(树套树版本已写，详见另一篇随笔)

记录一下水过去的代码：

View Code

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <cstdlib>
 6 
 7 #define lson l, m, rt << 1
 8 #define rson m + 1, r, rt << 1 | 1
 9 #define root 0, mx - 1, 1
10 
11 using namespace std;
12 
13 #define debug 0
14 
15 int min2(int _a, int _b){return _a < _b ? _a : _b;}
16 int max2(int _a, int _b){return _a > _b ? _a : _b;}
17 
18 const int maxn = 50001;
19 int tmp[maxn], s[maxn];
20 
21 int k_elem(int n, int *a, int k){//二分搜索第k大数
22     int t, key;
23     int i, j, l = 0, r = n - 1;
24     while (l < r){
25         for (key = a[((i = l - 1) + (j = r + 1)) >> 1]; i < j; ){
26             for (j--; key < a[j]; j--);
27             for (i++; a[i] < key; i++);
28             if (i < j) t = a[i], a[i] = a[j], a[j] = t;
29         }
30         if (k > j) l = j + 1;
31         else r = j;
32     }
33     return a[k];
34 }
35 
36 int main(){
37     int T, n, m;
38     int l, r, k;
39     char in[3];
40 
41     scanf("%d", &T);
42     while (T--){
43         scanf("%d%d", &n, &m);
44         for (int i = 1; i <= n; i++){
45             scanf("%d", &s[i]);
46         }
47         while (m-- && scanf("%s", in)){
48             if (in[0] == 'Q'){
49                 scanf("%d%d%d", &l, &r, &k);
50                 for (int i = l; i <= r; i++){
51                     tmp[i - l] = s[i];
52                 }
53                 printf("%d\n", k_elem(r - l + 1, tmp, k - 1));
54             }
55             else{
56                 scanf("%d%d", &l, &k);
57                 s[l] = k;
58             }
59         }
60     }
61 
62     return 0;
63 }

——written by Lyon

posted @ 2012-07-31 23:31 LyonLys 阅读(166) 评论(0) 收藏举报

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