POJ3784---Running Median(树状数组+二分)

Description
For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.

Input
The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M, (1 ≤ M ≤ 9999), giving the total number of signed integers to be processed. The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.

Output
For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.

Sample Input

3
1 9
1 2 3 4 5 6 7 8 9
2 9
9 8 7 6 5 4 3 2 1
3 23
23 41 13 22 -3 24 -31 -11 -8 -7
3 5 103 211 -311 -45 -67 -73 -81 -99
-33 24 56

Sample Output

1 5
1 2 3 4 5
2 5
9 8 7 6 5
3 12
23 23 22 22 13 3 5 5 3 -3
-7 -3

Source
Greater New York Regional 2009

把数据先离散化,然后遍历,把数字依次插入到树状数组里去, 求中值二分就行了

/*************************************************************************    > File Name: POJ3784.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年04月14日 星期二 11时57分18秒 ************************************************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace std;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;const int N = 12000;int tree[N];int xis[N];int arr[N];int st[N];int cnt;int n;int lowbit(int x){    return x & (-x);}void add(int x){    for (int i = x; i <= n; i += lowbit(i))    {        ++tree[i];    }}int sum(int x){    int ans = 0;    for (int i = x; i; i -= lowbit(i))    {        ans += tree[i];    }    return ans;}int search(int val){    int l = 1, r = cnt;    int mid;    while (l <= r)    {        mid = (l + r) >> 1;        if (xis[mid] == val)        {            break;        }        else if (xis[mid] > val)        {            r = mid - 1;        }        else        {            l = mid + 1;        }    }    return mid;}int main(){    int t;    scanf("%d", &t);    while (t--)    {        cnt = 0;        memset(tree, 0, sizeof(tree));        int icase;        scanf("%d%d", &icase, &n);        for (int i = 1; i <= n; ++i)        {            scanf("%d", &arr[i]);            xis[++cnt] = arr[i];        }        sort(xis + 1, xis + 1 + cnt);        cnt = unique(xis + 1, xis + 1 + cnt) - xis - 1;        int ret = 0;        int low = inf, high = -1;        for (int i = 1; i <= n; ++i)        {            int val = search(arr[i]);            add(val);            low = min(low, val);            high = max(high, val);            if (i == 1)            {                st[++ret] = xis[val];                continue;            }            if (i & 1)            {                int l = low, r = high, mid;                int a;                while (l <= r)                {                    mid = (l + r) >> 1;                    int num1 = sum(mid);                    if (num1 <= i / 2)                    {                        l = mid + 1;                    }                    else                    {                        r = mid - 1;                        a = mid;                    }                }                st[++ret] = xis[a];            }        }        printf("%d %d\n", icase, ret);        int g = 0;        for (int i = 1; i <= ret; ++i)        {            ++g;            printf("%d", st[i]);            if (i < ret && g < 10)            {                printf(" ");            }            else if (i < ret && g == 10)            {                g = 0;                printf("\n");            }            else            {                printf("\n");            }        }    }    return 0;}