hdu 4262(线段树)

题目:有一个圈,可以从某个位置取球,给出原有的顺序,有三种操作,左旋一次,右旋一次,取球,要求按顺序取球,问需要操作多少次

显然操作是确定的,每次将目标球旋转过来,找出左旋和右旋操作少的,然后取球。

 

每次旋转后相对距离不变,每次记录某个区间删掉了多少,就可以查询了。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 
 6 using namespace std;
 7 #define lson l, m, rt<<1
 8 #define rson m + 1, r, rt<<1|1
 9 typedef long long ll;
10 const int maxn = 1e5 + 5;
11 int n, pos[maxn], sum[maxn << 2];
12 ll ans;
13 void update(int p, int l, int r, int rt){
14     sum[rt]++;
15     if (l == r) return;
16     int m = (l + r) >> 1;
17     if (p <= m) update(p, lson);
18     else update(p, rson);
19 }
20 int query(int L, int R, int l, int r, int rt){
21     if (L > R) return 0;
22     if (L <= l && R >= r){
23         return sum[rt];
24     }
25     int m = (l + r) >> 1;
26     int ret = 0;
27     if (L <= m) ret += query(L, R, lson);
28     if (R > m) ret += query(L, R, rson);
29     return ret;
30 }
31 int main(){
32     int x;
33     while (~scanf("%d", &n) && n){
34         ans = 0;
35         memset(sum, 0, sizeof(sum));
36         for (int i = 0; i < n; ++i){
37             scanf("%d", &x);
38             pos[x] = i + 1;
39         }
40         int pre = 0, q, c, lx, rx;
41         for (int i = 1; i <= n; ++i){
42             x = pos[i];
43             if (x > pre){
44                 q = query(pre + 1, x, 1, n, 1);
45                 lx = x - pre - 1 - q;
46                 rx = n - i + 1 - lx;
47             }
48             else{
49                 q = query(x + 1, pre - 1, 1, n, 1);
50                 rx = pre - x - q;
51                 lx = n - i + 1 - rx;
52             }
53             //cout << " lx = "  << lx << " rx = " << rx << endl;
54             //int lx = x - 1 - query(1, x - 1, 1, n, 1);
55             //int rx = n - x + 1 - query(x + 1, n, 1, n, 1);
56             ans += min(lx, rx) + 1;
57             pre = x;
58             update(x, 1, n, 1);
59         }
60         printf("%lld\n", ans);
61     }
62     return 0;
63 }
View Code

 

posted @ 2013-11-05 16:00  Missa  阅读(341)  评论(0编辑  收藏  举报