线段树/树状数组 POJ 2182 Lost Cows

题目传送门

题意：n头牛，1~n的id给它们乱序编号，已知每头牛前面有多少头牛的编号是比它小的，求原来乱序的编号

分析：从后往前考虑，最后一头牛a[i] = 0，那么它的编号为第a[i] + 1编号：为1，倒数第二头牛的编号为除去最后一头牛的编号后的第a[i-1] + 1编号：为3，其他的类推，所以可以维护之前已经选掉的编号，求第k大的数字，sum[rt] 表示该区间已经被选掉的点的个数。另外树状数组也可以做，只不过用二分优化查找第k大的位置。

收获：逆向思维，求动态第K大

代码(线段树)：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/9 星期三 17:01:08
* File Name     :P.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 8e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct ST   {
    int sum[N<<2];
    void build(int l, int r, int rt)    {
        sum[rt] = 0;
        if (l == r) return ;
        int mid = (l + r) >> 1;
        build (lson);
        build (rson);
    }
    int query(int p, int l, int r, int rt)  {
        sum[rt]++;
        if (l == r) return l;
        int mid = (l + r) >> 1;
        int tmp = mid - l + 1 - sum[rt<<1];
        if (tmp >= p)    return query (p, lson);
        else    {
            return query (p - tmp, rson);
        }
    }
}st;
int a[N], ans[N];

int main(void)    {
    int n;
    while (scanf ("%d", &n) == 1)   {
        a[1] = 0;
        for (int i=2; i<=n; ++i) {
            scanf ("%d", &a[i]);
        }
        st.build (1, n, 1);
        for (int i=n; i>=1; --i)    {
            ans[i] = st.query (a[i] + 1, 1, n, 1);
        }
        for (int i=1; i<=n; ++i)    {
            printf ("%d\n", ans[i]);
        }
    }

    return 0;
}

代码(树状数组)：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/9 星期三 18:19:28
* File Name     :P_BIT.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 8e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n;
struct BIT  {
    int c[N];
    void init(void) {
        memset (c, 0, sizeof (c));
    }
    void updata(int i, int x)  {
        while (i <= n)   {
            c[i] += x;  i += i & (-i);
        }
    }
    int query(int i)    {
        int ret = 0;
        while (i)   {
            ret += c[i];    i -= i & (-i);
        }
        return ret;
    }
    int bsearch(int l, int r, int k)    {
        while (l <= r)  {
            int mid = (l + r) >> 1;
            if (mid - query (mid) >= k) r = mid - 1;
            else    l = mid + 1;
        }
        return l;
    }
}bit;
int a[N], ans[N];

int main(void)    {
    while (scanf ("%d", &n) == 1)   {
        a[1] = 0;
        for (int i=2; i<=n; ++i)    scanf ("%d", &a[i]);
        bit.init ();
        for (int i=n; i>=1; --i)    {
            ans[i] = bit.bsearch (1, n, a[i] + 1);
            bit.updata (ans[i], 1);
        }
        for (int i=1; i<=n; ++i)    {
            printf ("%d\n", ans[i]);
        }
    }

    return 0;
}