POJ-2299 Ultra-QuickSort (树状数组 离散 求逆序对数)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 64185		Accepted: 23972

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

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxn 500002
struct point{
int val, id;
bool operator < (const point& x){
return val < x.val;
}
}a[maxn];
int b[maxn], c[maxn];
inline int lowbit(int x){
return x & (-x);
}
void add(int x){
while(x < maxn){
c[x]++;
x += lowbit(x);
}
}
int getsum(int x){
int ans = 0;
while(x){
ans += c[x];
x -= lowbit(x);
}
return ans;
}
int main(){
int n, tot;
while(scanf("%d", &n) != EOF){
if(n == 0){
return 0;
}
a[0].val = -1;
for(int i = 1; i <= n; ++i){
scanf("%d", &a[i].val);
a[i].id = i;
}
sort(a + 1, a + 1 + n);
tot = 0;
for(int i = 1; i <= n; ++i){
if(a[i].val == a[i - 1].val){
b[a[i].id] = tot;
}
else{
b[a[i].id] = ++tot;
}
}
memset(c, 0, sizeof(c));
long long ans = 0;
for(int i = 1; i <= n; ++i){
ans += i - 1 - getsum(b[i]);
add(b[i]);
}
printf("%lld\n", ans);
}
}

/*
题意：求逆序对个数。

思路：树状数组。
*/