【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 =  + ;
 const int MAXM =  + ;
 //const int MAXM = 2000 + 10;
 const int MAXL = ;
 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 = ; i <= n; i++) {
          scanf("%d", &data[i]);
          C[i] = ;
          rem[i].val = data[i];
          rem[i].order = i;
      }
      sort(rem + , rem +  + n);
      for (int i = ; 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 = ;
    while (x > ){
          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 = ;
      //前面共 i - 1个数字
      for (int i = ; i <= n; i++){
          Ans += (i -  - 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 ;
 }