Minimum Inversion NumberTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13558Accepted Submission(s): 8277
Problem Description
The inversion number of a given number sequence a1, a2, …, an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.For a given sequence of numbers a1, a2, …, an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:a1, a2, …, an-1, an (where m = 0 – the initial seqence)a2, a3, …, an, a1 (where m = 1)a3, a4, …, an, a1, a2 (where m = 2)…an, a1, a2, …, an-1 (where m = n-1)You are asked to write a program to find the minimum inversion number out of the above sequences.
Input
The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.
Output
For each case, output the minimum inversion number on a single line.
Sample Input
101 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
CHEN, Gaoli
Source
线段树求逆序数 ！！！！！
以前只是听说过 ，现在完美解决
#include<iostream>#include<sstream>#include<algorithm>#include<cstdio>#include<string.h>#include<cctype>#include<string>#include<cmath>#include<vector>#include<stack>#include<queue>#include<map>#include<set>using namespace std;int sum=0;const int INF=5005;struct Tree{int left,right,count;} tree[5005<<2];int create(int root,int left,int right){tree[root].left=left;tree[root].right=right;if(left==right)return tree[root].count=0;int a,b,mid=(left+right)>>1;a=create(root<<1,left,mid);b=create(root<<1|1,mid+1,right);return tree[root].count=a+b;}int cal (int root,int left,int right){if(tree[root].left>right||tree[root].right<left)return 0;if(tree[root].left>=left&&tree[root].right<=right)return tree[root].count;int a,b,mid=(tree[root].left+tree[root].right)>>1;a=cal(root<<1,left,right);b=cal(root<<1|1,left,right);return a+b;}void update(int root,int pos,int val){if(tree[root].left==tree[root].right){tree[root].count=1;return ;}int mid=(tree[root].left+tree[root].right)>>1;if(pos<=mid)update(root<<1,pos,val);elseupdate(root<<1|1,pos ,val);tree[root].count=tree[root<<1].count+tree[root<<1|1].count;}int main(){int n,result=0 ,cnt[5004];while(scanf("%d",&n)!=EOF){result=0;create(1,0,n-1);for(int i=0; i<n; i++){scanf("%d",&cnt[i]);result+=cal(1,cnt[i],n-1);update(1,cnt[i],1);}int RR=result;for(int i=0; i<n; i++){result=result-cnt[i]+(n-cnt[i]-1);RR=min(RR,result);}printf("%d\n",RR);}return 0;}
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，同生天地间，为何我不能。
