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| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 44601 | Accepted: 16214 |
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 Input
Output
Sample Input
5 9 1 0 5 4 3 1 2 3 0
Sample Output
6 0
Source
分析：
求逆序对，因此利用归并排序，在排序过程中，记录逆序对的数目。
1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 #define max 500000 5 int c[max+5]; 6 int a[max+5]; 7 long long count; 8 void Merge(int *a,int first,int last){ 9 int p,q,temp,mid; 10 mid=(first+last)/2; 11 temp=first; 12 p=first; 13 q=mid+1; 14 while(p<=mid&&q<=last){ 15 if(a[p]>a[q]){ 16 count+=mid+1-p; 17 c[temp++]=a[q++]; 18 } 19 else{ 20 c[temp++]=a[p++]; 21 } 22 } 23 while(q<=last){ 24 c[temp++]=a[q++]; 25 } 26 q--; 27 while(p<=mid){ 28 //count+=q-mid; 29 c[temp++]=a[p++]; 30 } 31 int i=first; 32 for(;i<=last;i++){ 33 a[i]=c[i]; 34 } 35 } 36 void MergeSort(int *a,int first,int last){ 37 if(first==last){ 38 return; 39 } 40 int mid=(first+last)/2; 41 MergeSort(a,first,mid); 42 MergeSort(a,mid+1,last); 43 Merge(a,first,last); 44 } 45 int main(){ 46 int n; 47 while(scanf("%d",&n)&&n){ 48 count=0; 49 int i=0; 50 for(;i<n;i++){ 51 scanf("%d",&a[i]); 52 } 53 MergeSort(a,0,n-1); 54 cout<<count<<endl; 55 } 56 //cout<<1<<endl; 57 return 0; 58 }
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原文地址：http://www.cnblogs.com/Deribs4/p/4279953.html
