hdu 2838 Cow Sorting

题意：

给出一个数组，要求把他们排成升序，两个数字交换的代价是x + y，求代价的总和。

思路：

简单的逆序对变形，树状数组维护小于的cnt和sum即可。

代码：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <algorithm>
 4 using namespace std;
 5 typedef long long ll;
 6 const int N = 1e5 + 10;
 7 const int M = 100000;
 8 int a[N];
 9 int c[N];
10 ll sum[N];
11 int lowbit(int x)
12 {
13     return x&(-x);
14 }
15 void addcnt(int x)
16 {
17     for (int i = x;i <= M;i += lowbit(i)) c[i]++;
18 }
19 int getcnt(int x)
20 {
21     int ans = 0;
22     for (int i = x;i > 0;i -= lowbit(i)) ans += c[i];
23     return ans;
24 }
25 void addsum(int x,int y)
26 {
27     for (int i = x;i <= M;i += lowbit(i)) sum[i] += y;
28 }
29 ll getsum(int x)
30 {
31     ll ans = 0;
32     for (int i = x;i > 0;i -= lowbit(i)) ans += sum[i];
33     return ans;
34 }
35 int main()
36 {
37     int n;
38     while (scanf("%d",&n) != EOF)
39     {
40         memset(c,0,sizeof(c));
41         memset(sum,0,sizeof(sum));
42         for (int i = 0;i < n;i++)
43         {
44             scanf("%d",&a[i]);
45         }
46         ll tmp = 0;
47         ll ans = 0;
48         for (int i = 0;i < n;i++)
49         {
50             int cnt = getcnt(a[i]);
51             cnt = i - cnt;
52             ll s = getsum(a[i]);
53             s = tmp - s;
54             ans += s + 1LL * cnt * a[i];
55             addcnt(a[i]);
56             addsum(a[i],a[i]);
57             tmp += a[i];
58         }
59         printf("%lld\n",ans);
60     }
61     return 0;
62 }

 

posted @ 2018-05-24 04:55  qrfkickit  阅读(191)  评论(0)  编辑  收藏  举报