sgu 180 - Inversions （离散化+树状数组）

 - Inversions
Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u
Submit Status

Description

180. Inversions

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB
input: standard 
output: standard



There are N integers (1<=N<=65537) A1, A2,.. AN (0<=Ai<=10^9). You need to find amount of such pairs (i, j) that 1<=i<j<=N and A[i]>A[j].

Input
The first line of the input contains the number N. The second line contains N numbers A1...AN.

Output
Write amount of such pairs.

Sample test(s)

Input
 
 
5 
2 3 1 5 4
 
 

Output
 
 
3
 
 
一直wa 2
后来发现是没处理相同元素（我好傻逼啊。。。。）
离散化的时候，很重要的一项，当然是相同的元素，离散化的之后也要变成相同的。。。
上道题过了纯粹是数据水。。。
 
  1 /*************************************************************************
  2     > File Name: code/sgu/180.cpp
  3     > Author: 111qqz
  4     > Email: rkz2013@126.com 
  5     > Created Time: 2015年08月06日 星期四 16时40分53秒
  6  ************************************************************************/
  7 
  8 #include<iostream>
  9 #include<iomanip>
 10 #include<cstdio>
 11 #include<algorithm>
 12 #include<cmath>
 13 #include<cstring>
 14 #include<string>
 15 #include<map>
 16 #include<set>
 17 #include<queue>
 18 #include<vector>
 19 #include<stack>
 20 #define y0 abc111qqz
 21 #define y1 hust111qqz
 22 #define yn hez111qqz
 23 #define j1 cute111qqz
 24 #define tm crazy111qqz
 25 #define lr dying111qqz
 26 using namespace std;
 27 #define REP(i, n) for (int i=0;i<int(n);++i)  
 28 typedef long long LL;
 29 typedef unsigned long long ULL;
 30 const int inf = 0x7fffffff;
 31 const int N=7E4+7;
 32 struct Q
 33 {
 34     int val;
 35     int id;
 36 }q[N];
 37 int c[N];
 38 int ref[N];
 39 int n;
 40 bool cmp(Q a,Q b)
 41 {
 42     if (a.val<b.val)
 43     return true;
 44     return false;
 45 }
 46 
 47 int lowbit( int x)
 48 {
 49     return x&(-x);
 50 }
 51 void update( int x,int delta)
 52 {
 53     for ( int i = x; i < N ; i=i+lowbit(i) )
 54     {
 55     c[i] = c[i] + delta;
 56     }
 57 }
 58 int Sum( int x)
 59 {
 60     int res  =0;
 61     for ( int i = x; i >= 1 ; i = i-lowbit(i))
 62     {
 63     res = res + c[i];
 64     }
 65     return res;
 66 }
 67 int main()
 68 {
 69     while (scanf("%d",&n)!=EOF)
 70     {
 71 
 72       memset(c,0,sizeof(c));
 73       for ( int i = 1; i <= n ; i++ )
 74       {
 75         scanf("%d",&q[i].val);
 76     q[i].id  = i ;
 77     }
 78     sort(q+1,q+n+1,cmp);
 79     for ( int i = 1; i <= n ; i++ )
 80     {
 81     if (q[i].val!=q[i-1].val)
 82     {
 83         ref[q[i].id]=i;
 84     }
 85     else
 86     {
 87         ref[q[i].id]=ref[q[i-1].id];
 88     }
 89     }
 90 
 91   //  for ( int i = 1 ;i <= n ; i ++) cout<<ref[i]<<endl;
 92     LL  ans = 0;
 93     for ( int i = 1 ; i <= n ; i++ )
 94     {
 95     update(ref[i],1);
 96     ans = ans + i-Sum(ref[i]);
 97     }
 98     cout<<ans<<endl;
 99     }
100   
101     return 0;
102 }

 

 
posted @ 2015-08-06 17:09  111qqz  阅读(192)  评论(0)  编辑  收藏  举报