POJ 2299.Ultra-QuickSort

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题目:http://poj.org/problem?id=2299

AC代码(C++):

#include <iostream>#include <algorithm>#include <stdio.h>#include <vector>#include <queue>#include <math.h>#include <string>#include <string.h>#include <bitset>#define INF 0xfffffff#define MAXN 2005using namespace std;long long step;long long num[500005];long long temp[500005];void merge(int low, int mid, int high)  {      int i = low, j = mid + 1, k = low;      while (i <= mid && j <= high)      {          if (num[i] <= num[j])           {              temp[k++] = num[i++];          }          else           {              step += j - k;            temp[k++] = num[j++];          }      }      while (i <= mid) temp[k++] = num[i++];      while (j <= high) temp[k++] = num[j++];      for (i = low; i <= high; ++i)      {          num[i] = temp[i];      }  }    void mergeSort(int a, int b)  {      if (a < b)      {          int mid = (a + b) / 2;          mergeSort(a, mid);          mergeSort(mid + 1, b);          merge(a, mid, b);      }  }int main(){int n;while(cin>>n){if(n==0)break;step = 0;for(int i = 0; i < n; i++)scanf("%lld",&num[i]);mergeSort(0,n-1);cout<<step<<endl;}}

总结: 求交换次数, 其实就是求逆序数. 用归并排序可以快速求出逆序数. 注意这题数组大小可以大到50W, 则逆序数最大可以为n(n-1)/2=124999750000, 超过了int范围, 若答案用int表示就会WA, 所以用long long表示.

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