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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;
}
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原文地址：http://www.cnblogs.com/Running-Time/p/4795660.html
