归并排序求逆序数——HDU 4911

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Inversion

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

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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 12 2 13 02 2 1 

 

Sample Output

 12 

因为是相邻元素互换，故每换一次都会减少一对逆序数。所以直接愉快地套汝佳哥归并排序部分代码求出逆序数cnt，

如果cnt-k<0,则肯定能把原序列换成递增，即没有逆序数，否则输出cnt-k。

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;#define LL long long int shu[100005],vis[100005];LL cnt=0;void mer(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;mer(A,x,m,t);mer(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++];cnt+=m-p;}}for(i=x;i<y;i++){A[i]=t[i];}}}int main(){int n,k;while(scanf("%d%d",&n,&k)!=EOF){for(int i=1;i<=n;i++){scanf("%d",&shu[i]);}cnt=0;mer(shu,1,n+1,vis);if(cnt-k<0) printf("%d\n",0);else  printf("%I64d\n",cnt-k);}}