hust 1347 - Reverse Number

题目描述

Given a non-negative integer sequence A with length N, you can exchange two adjacent numbers each time. After K exchanging operations, what’s the minimum reverse number the sequence can achieve? The reverse number of a sequence is the number of pairs (i, j) such that i < j and Ai > Aj

输

入There are multiple cases. For each case, first line contains two numbers: N, K 2<=N<=100000, 0 <= K < 2^60 Second line contains N non-negative numbers, each of which not greater than 2^30

输出

Minimum reverse number you can get after K exchanging operations.

样例输入

3 1
3 2 1
5 2
5 1 4 3 2

样例输出

Case #1: 2
Case #2: 5

本题当然是先求逆序对，归并排序，树状数组，线段树，什么都可以，主要是先求出来，做k次变化，我们的目标不就是找刚好是逆序的变换，这样再看看k是否大于逆序对数的情况，不就完了

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define  inf 0x0f0f0f0f

using namespace std;
const int maxn=100000+10;

int a[maxn],b[maxn],ans;

void merge_sort(int* A,int x,int y,int*T)
{
     if (y-x>1)
     {
          int m=x+(y-x)/2;
          int p=x, q=m, i=x;
          merge_sort(A,x,m,T);
          merge_sort(A,m,y,T);
          while (p<m || q<y)
          {
               if (q>=y || (p<m && A[p]<=A[q])) T[i++]=A[p++];
               else {T[i++]=A[q++];ans+=m-p;}
          }
          for (int i=x;i<y;i++) A[i]=T[i];
     }
}

int main()
{
     int n,k,t=0;
     while(scanf("%d%d",&n,&k)!=EOF)
     {
          t++;
          for (int i=0;i<n;i++) scanf("%d",&a[i]);
          ans=0;
          merge_sort(a,0,n,b);
          ans-=k;
          bool cut=false;
          for (int i=0;i<n-1;i++) if (a[i]==a[i+1]) {cut=true;break;}
          if (ans<0)
          {
               if ((-ans)%2==0) ans=0;
               else
               {
                    if(cut) ans=0;
                    else ans=1;
               }
          }
          printf("Case #%d: %d\n",t,ans);
     }
     return 0;
}

作者 chensunrise

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