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HDU 4911

时间:2015-08-09 20:18:38      阅读:93      评论:0      收藏:0      [点我收藏+]

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Description

bobo has a sequence a 1,a 2,…,a n. He is allowed to swap two adjacent numbers for no more than k times. 
Find the minimum number of inversions after his swaps. 
Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and a i>a j.
 

Input

The input consists of several tests. For each tests: 
The first line contains 2 integers n,k (1≤n≤10 5,0≤k≤10 9). The second line contains n integers a 1,a 2,…,a n (0≤a i≤10 9).
 

Output

For each tests: 
A single integer denotes the minimum number of inversions.
 

Sample Input

3 1 2 2 1 3 0 2 2 1
 

Sample Output

1 2
 
 
题意:给一个序列和交换次数k,求交换后序列最小的的逆序数....
 
 
解题思路:先求出原始序列的逆序数,然后再减去交换次数,就是答案...值得注意的是,如果交换次数减原始逆序数小于0则,答案为0...
 
 
求逆序数,这里用到了归并排序..
 
 
 
 
代码如下:0
 
 
 
 1 #include <stdio.h>
 2 #include <string.h>
 3 int n,k,A[100000],T[100000];
 4 long long ans;
 5 void guibing(int* A,int x,int y,int* T)
 6 {
 7     if(y-x>1){
 8         int m=x+(y-x)/2;
 9         int p=x,q=m,i=x;
10         guibing(A,x,m,T);
11         guibing(A,m,y,T);
12         while(p<m||q<y){
13             if(q>=y||(p<m&&A[p]<=A[q])) T[i++]=A[p++];
14             else {
15                 T[i++]=A[q++];
16                 ans+=m-p;
17             }
18         }
19         for(i=x;i<y;i++) A[i]=T[i];
20     }
21 }
22 int main()
23 {
24     while(scanf("%d%d",&n,&k)==2&&n){
25         ans=0;
26         for(int i=0;i<n;i++){
27             scanf("%d",&A[i]);
28         }
29         memset(T,0,sizeof(T));
30         guibing(A,0,n,T);
31         if(ans-k<0)
32             ans=0;
33         else
34             ans=ans-k;
35         printf("%I64d\n",ans);
36     }
37     return 0;
38 }

 

 
 

HDU 4911

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原文地址:http://www.cnblogs.com/huangguodong/p/4716033.html

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