1394 ( Minimum Inversion Number ) (线段树)

http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16259    Accepted Submission(s): 9891


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
 
线段树水题,输入一个x,查询【x,n-1】区间有多少数,然后update【x】,如果先更新,那么查询的值就是【x+1,n-1】了,可能会出错

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
const int SIZE=5e3+10;
const int maxn=500000;
const int Mod=1e8;
int sum[SIZE<<2];
int a[SIZE];
void pushup(int rt){
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt){
    memset(sum,0,sizeof(sum));
}
void update(int x,int l,int r,int rt){
    if(l==r){
        sum[rt]=1;
        return;
    }
    int m=(l+r)>>1;
    if(x<=m)update(x,lson);
    else update(x,rson);
    pushup(rt);
}
int query(int x,int l,int r,int rt){
    if(x<=l){
        return sum[rt];
    }
    int m=(l+r)>>1;
    int cnt=0;
    if(x


你可能感兴趣的:(线段树)