Hduoj1394【线段树+逆序数处理】
/*Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11841 Accepted Submission(s): 7259
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
16
Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
Recommend
Ignatius.L
*/
#include<stdio.h>
int segtree[20010], fa[20010], a[5010];
void build(int i, int x, int y)
{
segtree[i] = 0;//初始化个数为0
if(x == y)
{
fa[x] = i;//记录原始值所在二叉树的位置
return ;
}
int j = (x+y)>>1;
build(i<<1, x, j);
build(i<<1|1, j + 1, y);
}
void update(int i)//从叶子开始往上更新
{
if(i == 1)//更新到根节点结束
return;
int j = i>>1;//当前节点的父节点
segtree[j] = segtree[j<<1] + segtree[j<<1|1]; //更新父节点的值
update(j);//递归更新
}
int query(int i, int l, int r, int L, int R)//返回l到r中以出现的个数
{
if(l <= L && R <= r)//如果查询区间包含当前区间，返回该区间的值
return segtree[i];
int num = 0, j = (L+R)>>1;
if(l <= j)//否则继续缩小当前区间
num += query(i<<1, l, r, L, j);
if(r > j)
num += query(i<<1|1, l, r, j+1, R);
return num;//最终返回查询区间的值
}
int main()
{
int i, j, k, n, m;
while(scanf("%d", &n) != EOF)
{
m = 0;
build(1, 0, n-1);//注意这里是从0开始的~~
for(i = 0; i < n; i++)
{
scanf("%d", &a[i]);
/*必须先查询后更新 */
/*我这里是将线段树中输入值的位置标记为1*/
/*如果先更新后查询每次会多加一个本身的个数*/
/*其次查询最好是查询后半段，而不是查询前半段再拿a【i】去减去查询的值 */
m += query(1, a[i], n-1, 0, n-1);//从根区间开始查询当前值之前已出现的数字个数
segtree[fa[a[i]]] = 1;//二叉树对应数字的位置标为1
update(fa[a[i]]); //向上更新二叉树的区间和
}
int min = m;
/*
因为序列为[0, n-1]，若最前面一个数为x，序列中比x
小的数为[0, x-1], 共x个，比x大的数为[x+1, n-1],
共n-x-1个，将x移到最后，比x小的数的逆序数均减1，
x的前面比x大的数有n-x-1个，x的逆序数增加n-x-1。
所以新序列的逆序数为原序列的逆序数加上n-2*x-1。
*/
for(i = 0; i < n; i ++)
{
m = m - a[i]*2 + n-1;
if( min > m )
min = m;
}
printf("%d\n", min);
}
return 0;
}
题意：给定一个序列，求序列的逆序数对数，其次该序列的前n位可以移到最后产生一个新的序列，总共有n种序列，求这n个序列中逆序数对数最少的个数。
思路：这题有两个关键，第一是想到原输入序列的逆序数对数和用线段树去求出来，其思路就是对于每一个输入的数都标记进入二叉树，其次二叉树区间m~n的意思就是在序列m~n中已经出现了几个数，然后每次输入一个数就可查询“输入的数到n-1“之间的数已经出现了几个，这个值就是该输入的数所对应的逆序数的对数。
第二个关键就是，求除了输入的原序列以外其余n-1种序列的逆序数对数的和，这个我就不说了上面注释有。
体会：要学会活用线段树是个难点。