Ultra-QuickSort(树状数组求逆序对数)

Ultra-QuickSort
题目链接：http://poj.org/problem?id=2299
Time Limit: 7000MS   Memory Limit: 65536K
Total Submissions: 51641   Accepted: 18948

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,

Ultra-QuickSort produces the output 
0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0
题解：树状数组求逆序对数，就是求对于每一个数后面有多少个数字比它自己的数字小，那么从后向前遍历所有的数字，a[x]表示的是当前状态下小于等于x的值得个数.那么从后往前的扫描所有的数值统计后，再把这个数字对应的a[x]++;这样在扫描它前面的数的时候就相当于考虑这个数了，这种总是求a的前缀和和对单独点修改的操作可以使用树状数组解决。
注意这个题要解决的数很大，所以要离散化一下，这里介绍两种离散化的方法：
1.写一个二分查找的函数，先sort()一下，然后对于每个值，find(x)返回的下标值就是离散化后的结果，这里注意因为树状数组不能处理下标是0的情况，所以要将编号从1开始。而且要注意结果有可能会超int所以要用long long ,因为这个wa了好多次。
代码：
 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 #define ll long long
 6 const ll N = 500005;
 7 ll a[N];
 8 ll mp[N];
 9 ll tree[N];
10 ll lowbit(ll x){
11     return x&(-x);
12 }
13 ll sum(ll x){
14     ll ans = 0;
15     while(x > 0){
16         ans += tree[x];
17         x-=lowbit(x);
18     }
19     return ans;
20 }
21 void add(ll x){
22     while(x<=N){
23         tree[x]++;
24         x+=lowbit(x);
25     }
26 }
27 ll n;
28 ll find(ll x){
29     ll l = 0;
30     ll r = n-1;
31     ll mid = (l+r)/2;
32     while(l<=r){
33         if(a[mid]==x) return mid;
34         else if(a[mid]<x) l = mid+1;
35         else if(a[mid]>x) r = mid-1;
36         mid = (l+r)/2;
37     }
38 }
39 int main()
40 {
41     while(~scanf("%d",&n))
42     {
43         if(n==0) return 0;
44         memset(tree,0,sizeof(tree));
45         for(ll i = 0; i < n; i++){
46             scanf("%I64d",&mp[i]);
47             a[i] = mp[i];
48         }
49         sort(a,a+n);
50         ll ans = 0;
51         for(ll i = n-1; i >= 0; i--){//从后往前扫描
52             mp[i] = find(mp[i])+1;
53             //prllf("%d \n",mp[i]);
54             ans += sum(mp[i]-1);
55             //prllf("ans = %d ",ans);
56             add(mp[i]);
57         }
58         printf("%I64d\n",ans);
59     }
60     return 0;
61 }

 

 

posted on 2016-02-25 15:25  若流芳千古  阅读(285)  评论(0)  编辑  收藏  举报

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