hdu 1394 Minimum Inversion Number -求逆序对的个数- 线段树单节点更新

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19711    Accepted Submission(s): 11845

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

ZOJ Monthly, January 2003

 

题意

n个数分别是 0~n-1 ,给出这n个数的一个序列，可以做如题的转换， 问转换后最少有多少个逆序对

解题思路

用线段树维护当前区间出现过几个数， 插入a[i] 之前插入的比 a[i] 大的数的个数，即为 a[i]构成的逆序对的个数

先处理出初始序列的逆序对个数，然后从头枚举将a[i] 放 到最后，逆序对个数会变为几 。

a[i]一开始在第一个位置，比他小的肯定都在其后面，产生a[i]个逆序对
 所以将a[i]放到最后需 要
 加上多产生的(n-a[i]-1)个逆序对

再减去a[i]放最后之前构成逆序对的个数

代码

#include <iostream>#include <cstdio>#include <map>#include <cmath>#include <string.h>#include <algorithm>#include <set>#include <sstream>#include <vector>#include <queue>#include <stack>using namespace std;const int maxn = 5000*4;const int INF = 0x3f3f3f3f;struct node{    int l,r;    int sum;///区间l-r有几个数了}tree[maxn];void buildtree(int node,int b,int e){    int mid = (b+e)/2;    tree[node].l = b;    tree[node].r  = e;    tree[node].sum = 0;    if(b==e) return;    if(b <= mid) buildtree(node*2,b,mid);    if(e > mid) buildtree(node*2+1,mid+1,e);}void update(int node,int ql,int qr){    if(ql <= tree[node].l && qr >= tree[node].r){        tree[node].sum ++;        return;    }    int mid  = (tree[node].l + tree[node].r)/2;    if(ql <= mid) update(node*2,ql,qr);    if(qr > mid) update(node*2+1,ql,qr);    tree[node].sum = tree[node*2].sum + tree[node*2+1].sum;}int query(int node,int ql,int qr){    int  ans1 = 0,ans2 = 0;    if(ql <= tree[node].l && qr >= tree[node].r)        return tree[node].sum;    int mid = (tree[node].l + tree[node].r)/2;    if(ql <= mid) ans1 = query(node*2,ql,qr);    if(qr > mid) ans2 = query(node*2+1,ql,qr);    return ans1+ans2;}int main(){    int n;    int a[maxn];    while(scanf("%d",&n) != EOF){        int sum = 0;///初始情况下，逆序对的个数        buildtree(1,0,n-1);        for(int i =0;i<n;++i){            scanf("%d",&a[i]);            sum += query(1,a[i],n-1);///在他之前插入的比他大的数有几个            update(1,a[i],a[i]);        }        int res = sum;        for(int i = 0; i < n; ++i){///将a[i]放到后面            ///a[i]一开始在第一个位置，比他小的肯定都在其后面，产生a[i]个逆序对            ///所以将a[i]放到最后需要            ///加上多产生的(n-a[i]-1)个逆序对再减去a[i]放最后之前构成逆序对的个数            sum += (n-a[i]-1) - a[i];            res = min(res,sum);        }        printf("%d\n",res);    }    return 0;}