HDU1394 Minimum Inversion Number【线段树 点修改+区间查询】
Minimum Inversion Number
http://acm.hdu.edu.cn/showproblem.php?pid=1394
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26274 Accepted Submission(s): 15429
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
Sample Output
16Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
题意
有一个n个整数的排列，这n个整数就是0,1,2,3...n-1这n个数(但不一定按这个顺序给出)。现在先计算一下初始排列的逆序数，然后把第一个元素a1放到an后面去，形成新排列a2 a3 a4...an a1，然后再求这个排列的逆序数。继续执行类似操作(一共要执行n-1次)直到产生排列an a1 a2...an-1为止。计算上述所有排列的逆序数，输出最小逆序数。
思路
为了方便处理，我们统一将输入的数加一，这样数的范围就是[1,n]。设ans表示初始数列的逆序数，a[i]表示当前数列的第一个数，那么将a[i]移动到最后形成的新的数列的逆序数为多少？是不是ans2=ans-(a[i]-1)+(n-a[i])=ans+n+1-2*a[i]，很容易想到，就是在原来的逆序数的基础上进行加减，减a[i]-1是因为把a[i]移动到最后了，原本和a[i]构成逆序对的(x,a[i])，x
那么现在问题的关键就是求出初始数列的逆序数，此处使用线段树进行求解，设结点rt表示的区间是[l,r]，那么结点rt存放的信息为[l,r]范围内的数出现的次数，我们可以这样求初始数列的逆序数，求出每个数前面比它大的数的数量，具体看代码
C++代码
#include
#include
#include
using namespace std;
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
const int N=5010;
int a[N],sum[N<<2];
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
//单点修改 sum[L]+=1
void update(int L,int l,int r,int rt)
{
if(l==r)
{
sum[rt]+=1;
return;
}
int m=(l+r)>>1;
if(L<=m)
update(L,ls);
else
update(L,rs);
pushup(rt);
}
//区间查询,查询当前线段树中值的范围在[L,R]中的数的数量
int query(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
return sum[rt];
int m=(l+r)>>1;
int res=0;
if(L<=m)
res+=query(L,R,ls);
if(R>m)
res+=query(L,R,rs);
return res;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
//由于建树是让sum[i]=0，因此直接将sum数组置0就能代替建树过程
memset(sum,0,sizeof(sum));
int ans=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]++;//让a[i]加一，使得a[i]在[1,n]的范围而不是原来的[0,n-1]，这样更好处理
//查询a[i]的前面有多少数比a[i]大,即查询多又是个元素在[a[i]+1,n]的范围内
if(a[i]