【POJ2266】【树状数组+离散化】Ultra-QuickSort

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

【分析】

开始离散化用MAP T了半天，不活了..

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <utility>
#include <iomanip>
#include <string>
#include <cmath>
#include <map>

const int MAXN = 500000 + 10;
const int MAXM = 500000 + 10;
//const int MAXM = 2000 + 10;
const int MAXL = 10;
using namespace std;
struct DATA{
       int val;
       int order;
       bool operator < (DATA b)const{
            return val < b.val;
       }
}rem[MAXN];
typedef long long ll;
int n;
int data[MAXN];
int C[MAXN];

void init(){
     for (int i = 1; i <= n; i++) {
         scanf("%d", &data[i]);
         C[i] = 0;
         rem[i].val = data[i];
         rem[i].order = i;
     }
     sort(rem + 1, rem + 1 + n);
     for (int i = 1; i <= n; i++) data[rem[i].order] = i;
     //for (int i = 1; i <= n; i++) printf("%d\n", data[i]);printf("\n");
}
int lowbit(int x){return x&-x;}
int sum(int x){
   int cnt = 0;
   while (x > 0){
         cnt += C[x];
         x -= lowbit(x);
   }
   return cnt;
}
void add(int x){
   while (x <= n){
         C[x]++;
         x += lowbit(x);
   }
   return;
}

void work(){
     ll Ans = 0;
     //前面共 i - 1个数字 
     for (int i = 1; i <= n; i++){
         Ans += (i - 1 - sum(data[i]));//严格大于
         add(data[i]); 
     }
     printf("%lld\n", Ans);
}

int main(){
     #ifdef LOCAL
     freopen("data.txt", "r", stdin);
     freopen("out.txt", "w", stdout); 
     #endif 
     while (scanf("%d", &n) && n){
           init();
           work();
     }
     return 0;
}