HDU 1394 (树状数组 & 线段树 两种做法)

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16165    Accepted Submission(s): 9836

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

 

题目意思：求逆序数，然后将第一个数放到最后，求出逆序数，再将新的数组的第一个数放到最后，在求出新的逆序数，最后求这些逆序数中的最小的一个

首先用了线段树写的，首先将数组插入到线段树种，然后对从这个数到最后进行查询，就可以找出在插入的这个数之前的大于这个数的个数，这道题是按照找前面的比他大的数来累加的，和从后面找比他小的个数的和是一样的

#include <iostream>#include <cstring>#include <cstdio>#define Max 5005#define ls l,m,rt<<1#define rs m+1,r,rt<<1|1using namespace std;int segTree[Max<<2];//存的是树的下面的最小的值int ary[Max];//存的是输入的数字int vis[Max<<2];//inline max(int a,int b){return a>b?a:b;}inline min(int a,int b){return a>b?b:a;}void push_up(int rt){segTree[rt] = segTree[rt<<1] + segTree[rt<<1|1];}void build(int l,int r,int rt)//构建线段树，并求出每个节点的下面分支的最小值{segTree[rt] = 0;if(l == r)return ;int m = (l+r)>>1;build(ls);build(rs);push_up(rt);}void update(int site,int l,int r,int rt)//单点更新{if(l == r){segTree[rt]++;return ;}int m = (l + r) >> 1;if(site <= m)update(site,ls);elseupdate(site,rs);push_up(rt);}int query(int ll,int rr,int l,int r,int rt){if(l >= ll && r <= rr)return segTree[rt];int m = (l + r) >> 1;int ans =0 ;if(ll <= m)ans += query(ll,rr,ls);if(rr > m)ans += query(ll,rr,rs);return ans;}int main(){    int n,m;int i,j,k;int x[Max];    while(cin>>n){build(0,n-1,1);int sum = 0;for(i=0;i<n;i++){scanf("%d",&x[i]);sum += query(x[i],n-1,0,n-1,1);update(x[i],0,n-1,1);}int res = sum;for(i=0;i<n;i++){sum += n-x[i] - x[i] - 1;res = min(sum,res);}cout<<res<<endl;    }      return 0; }

然后试着用树状数组也写了一下，原理和线段树是相同的，只不过树状数组找比他前面的大的数，是先找比他小的数，然后再拿总的数减去比他小的数就可以求出比他大的数了

#include <iostream>#include <cstring>#include <cstdio>#define Max 5005using namespace std;int c[Max],nn;inline max(int a,int b){return a>b?a:b;}inline min(int a,int b){return a>b?b:a;}int lowbit(int k){return k&(-k);}void add(int k){while(k <= nn){c[k]++;k += lowbit(k);}}int sum(int k){int s = 0;while(k){s += c[k];k -= lowbit(k);}return s;}int main(){    int n,m;int i,j,k;int x[Max];    while(cin>>nn){m=0;memset(c,0,sizeof(c));for(i=0;i<nn;i++){scanf("%d",&x[i]);m += (sum(nn) - sum(x[i]+1));add(x[i]+1);}int res = m;for(i=0;i<nn;i++)        {m += nn - x[i] - x[i] -1;res = min(res,m);}cout<<res<<endl;    }      return 0; }

最终发现树状数组还是效率更高一点点