树状数组_二分_POJ-2182

题目链接：http://poj.org/problem?id=2182

这题似乎可以有多解，但题目里似乎没有specialjudge

本题解法较为简单，注意到每次可以确定队尾的元素，在剩下的未确定的元素中排第几。

至于值为多少就可以用二分和树状数组来解决。

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 #define lowbit(a) ((a)&(-a))
 7 #define max(a, b) ((a)>(b)?(a):(b))
 8 #define min(a, b) ((a)<(b)?(a):(b))
 9 #define MAXN 10010
10 #define PI 3.1415926
11 #define E 2.718281828
12 #define INF 0x777777f
13 typedef long long LL;
14
15 struct BIT{
16     #define bMAXN 10010
17     int bCnt, sum[bMAXN];
18     void add(int p, int v){
19         for ( ; p <= bCnt; p += lowbit(p))
20             sum[p] += v;
21     }
22     int query(int p){
23         int ret = 0;
24         for ( ; p; p -= lowbit(p))
25             ret += sum[p];
26         return ret;
27     }
28 } t;
29 int n, a[MAXN], ans[MAXN];
30
31 int find(int v){
32     int l = 1, r = n, mid;
33     while (l <= r){
34         mid = (l+r)>>1;
35         int tmp = t.query(mid);
36         if (tmp < v) l = mid+1;
37         if (tmp >= v) r = mid-1;    //每次找到最左边的那个，防止访问已被删除的节点
38     } return l;
39 }
40 int main(){
41     scanf("%d", &n); t.bCnt = n;
42     for (int i = 2; i <= n; i++) scanf("%d", &a[i]);
43     for (int i = 1; i <= n; i++) t.add(i, 1);    //初始化
44     for (int i = n; i >= 1; i--){
45         ans[i] = find(a[i]+1);
46         t.add(ans[i], -1);
47     }
48     for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
49     return 0;
50 }