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  • Minimum Inversion Number(线段树求逆序数)

    Minimum Inversion Number

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15517    Accepted Submission(s): 9467

    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
    10 1 3 6 9 0 8 5 7 4 2
    mergesort
     
    Sample Output
    16
     
    Author
    CHEN, Gaoli

    题解:线段树;也可以用归并排序,也可以用树状数组;注意每次把第一个放在最后这个条件;归并:http://www.cnblogs.com/handsomecui/p/4814442.html

    mergesort

    代码:

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    #include<map>
    using namespace std;
    const int MAXN=5010;
    #define lson root<<1,l,mid
    #define rson root<<1|1,mid+1,r
    #define pushup tree[root]=tree[root<<1]+tree[root<<1|1]
    #define mem(x,y) memset(x,y,sizeof(x))
    int tree[MAXN<<2];
    int a[MAXN];
    int ans; 
    void update(int v,int root,int l,int r){
    	int mid=(l+r)>>1;
    	if(l==v&&r==v){
    		tree[root]++;
    		return;
    	}
    	if(mid>=v)update(v,lson);
    	if(mid<v)update(v,rson);
    	pushup;
    }
    void query(int L,int R,int root,int l,int r){
    	int mid=(l+r)>>1;
    	if(l>=L&&r<=R){
    		ans+=tree[root];
    		return;
    	}
    	if(mid>=L)query(L,R,lson);
    	if(mid<R)query(L,R,rson);
    }
    int main(){
    	int N;
    	while(~scanf("%d",&N)){
    		mem(tree,0);
    		ans=0;
    		int x;
    		for(int i=0;i<N;i++){
    			scanf("%d",&x);a[i]=x;
    			query(x+1,N-1,1,0,N-1);
    		//	printf("%d
    ",ans);
    			update(x,1,0,N-1);
    		}
    		int cnt=ans;
    		for(int i=0;i<N;i++){
    			cnt=cnt+N-1-a[i]-a[i];
    			ans=min(ans,cnt);
    		}
    		printf("%d
    ",ans);
    	}
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/handsomecui/p/5020297.html
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