Ultra-QuickSort
| 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 Ultra-QuickSort produces the output
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 ",mp[i]); 54 ans += sum(mp[i]-1); 55 //prllf("ans = %d ",ans); 56 add(mp[i]); 57 } 58 printf("%I64d ",ans); 59 } 60 return 0; 61 }