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：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define root 0, mx - 1, 1

using namespace std;

#define debug 0

int min2(int _a, int _b){return _a < _b ? _a : _b;}
int max2(int _a, int _b){return _a > _b ? _a : _b;}

const int maxn = 50001;
int tmp[maxn], s[maxn];

int k_elem(int n, int *a, int k){//二分搜索第k大数
    int t, key;
    int i, j, l = 0, r = n - 1;
    while (l < r){
        for (key = a[((i = l - 1) + (j = r + 1)) >> 1]; i < j; ){
            for (j--; key < a[j]; j--);
            for (i++; a[i] < key; i++);
            if (i < j) t = a[i], a[i] = a[j], a[j] = t;
        }
        if (k > j) l = j + 1;
        else r = j;
    }
    return a[k];
}

int main(){
    int T, n, m;
    int l, r, k;
    char in[3];

    scanf("%d", &T);
    while (T--){
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++){
            scanf("%d", &s[i]);
        }
        while (m-- && scanf("%s", in)){
            if (in[0] == 'Q'){
                scanf("%d%d%d", &l, &r, &k);
                for (int i = l; i <= r; i++){
                    tmp[i - l] = s[i];
                }
                printf("%d\n", k_elem(r - l + 1, tmp, k - 1));
            }
            else{
                scanf("%d%d", &l, &k);
                s[l] = k;
            }
        }
    }

    return 0;
}

——written by Lyon