Frosh Week
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1545 Accepted Submission(s): 497
Problem Description
During Frosh Week, students play various fun games to get to know each other and compete against other teams. In one such game, all the frosh on a team stand in a line, and are then asked to arrange themselves according to some criterion, such as their height, their birth date, or their student number. This rearrangement of the line must be accomplished only by successively swapping pairs of consecutive students. The team that finishes fastest wins. Thus, in order to win, you would like to minimize the number of swaps required.
Input
The first line of input contains one positive integer n, the number of students on the team, which will be no more than one million. The following n lines each contain one integer, the student number of each student on the team. No student number will appear more than once.
Output
Output a line containing the minimum number of swaps required to arrange the students in increasing order by student number.
Sample Input
3 3 1 2
Sample Output
2
Source
1 /* 2 树状数组求逆序数 3 */ 4 #include<stdio.h> 5 #include<string.h> 6 #include<stdlib.h> 7 #define maxn 1000000 8 int nn; 9 __int64 tol; 10 int aa[maxn+5]; 11 12 struct node 13 { 14 int id; 15 int val; 16 }stu[maxn+5]; 17 //低位操作 18 int lowbit(int x) 19 { 20 return x&(-x); 21 } 22 23 void ope(int x) 24 { 25 while(x<=nn) 26 { 27 aa[x]++; 28 x+=lowbit(x); 29 } 30 } 31 32 __int64 sum(int x) 33 { 34 __int64 ans=0; 35 while(x>0) 36 { 37 ans+=aa[x]; 38 x-=lowbit(x); 39 } 40 return ans; 41 } 42 int cmp(void const *a,void const *b) 43 { 44 return (*(struct node *)a).val - (*(struct node *)b).val; 45 } 46 int main() 47 { 48 int i,val; 49 while( scanf("%d",&nn)!=EOF) 50 { 51 tol=0; 52 memset(aa,0,sizeof(int)*(nn+5)); 53 for(i=0;i<nn;i++) 54 { 55 scanf("%d",&stu[i].val); 56 stu[i].id=i+1; 57 } 58 qsort(stu,nn,sizeof(struct node),cmp); 59 for(i=0;i<nn;i++) 60 { 61 tol+=sum(nn)-sum(stu[i].id); 62 ope(stu[i].id); 63 } 64 printf("%I64d ",tol); 65 } 66 return 0; 67 }
运用归并排序求解:
递归版
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #define maxn 1000000 5 int aa[maxn+5]; 6 int cc[maxn+5]; 7 __int64 tol; 8 void merge(int low ,int mid ,int hight) 9 { 10 int i,j,k; 11 i=low; 12 j=mid; 13 k=0; 14 while(i<mid&&j<hight) 15 { 16 if(aa[i]>aa[j]) 17 { 18 cc[k++]=aa[j++]; 19 tol+=mid-i; 20 } 21 else 22 cc[k++]=aa[i++]; 23 } 24 for( ; i<mid; i++) 25 cc[k++]=aa[i]; 26 for( ; j<hight ; j++ ) 27 cc[k++]=aa[j]; 28 k=0; 29 for(i=low;i<hight;i++) 30 aa[i]=cc[k++]; 31 } 32 void merge_sort(int st ,int en) 33 { 34 int mid; 35 if(st+1<en) 36 { 37 mid=st+(en-st)/2; 38 merge_sort(st,mid); 39 merge_sort(mid,en); 40 merge(st,mid,en); 41 } 42 } 43 44 int main() 45 { 46 int n,i; 47 while(scanf("%d",&n)!=EOF) 48 { 49 tol=0; 50 for(i=0;i<n;i++) 51 scanf("%d",aa+i); 52 merge_sort(0,n); 53 printf("%I64d ",tol); 54 } 55 return 0; 56 }
非递归版的归并排序
代码：
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #define maxn 1000000 5 int aa[maxn+5]; 6 int cc[maxn+5]; 7 __int64 tol; 8 void merge(int low ,int mid ,int hight) 9 { 10 int i,j,k; 11 i=low; 12 j=mid; 13 k=0; 14 while(i<mid&&j<hight) 15 { 16 if(aa[i]>aa[j]) 17 { 18 cc[k++]=aa[j++]; 19 tol+=mid-i; 20 } 21 else 22 cc[k++]=aa[i++]; 23 } 24 for( ; i<mid; i++) 25 cc[k++]=aa[i]; 26 for( ; j<hight ; j++ ) 27 cc[k++]=aa[j]; 28 k=0; 29 for(i=low;i<hight;i++) 30 aa[i]=cc[k++]; 31 } 32 void merge_sort(int st,int en) 33 { 34 int s,t,i; 35 t=1; 36 while(t<=(en-st)) 37 { 38 s=t; 39 t=2*t; 40 i=st; 41 while(t+i<=en) 42 { 43 merge(i,i+s,i+t); 44 i+=t; 45 } 46 if(i+s<en) 47 merge(i,i+s,en); 48 } 49 if(st+s<en) 50 merge(st,st+s,en); 51 52 } 53 54 int main() 55 { 56 int n,i; 57 while(scanf("%d",&n)!=EOF) 58 { 59 tol=0; 60 for(i=0;i<n;i++) 61 scanf("%d",aa+i); 62 merge_sort(0,n); 63 printf("%I64d ",tol); 64 } 65 return 0; 66 }