HDU--杭电--Minimum Inversion Number--线段树--基础

Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8314    Accepted Submission(s): 5106
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

题意就是给你一个排列，求出逆序数，然后根据每次把第一个放最后的方式得到新排列再求逆序数，最后输出最小的
线段树的应用就是求逆序数的时候，记录每个数字是否出现以及个数
#include <iostream>#include <cstdio>using namespace std;struct node{    int l,r,n;}tree[1000000];void setit(int l,int r,int step){    tree[step].l=l;    tree[step].r=r;    tree[step].n=0;    if(l==r)return;    setit(l,(l+r)/2,2*step);    setit((l+r)/2+1,r,2*step+1);}void insert(int l,int r,int k,int step)//插入一个k{    tree[step].n++;    if(l==r)return;    int mid=(l+r)/2;    if(mid>=k)insert(l,mid,k,2*step);    else insert(mid+1,r,k,2*step+1);}int find(int l,int r,int k,int step)//查找并返回所有小于k的数的个数{    int mid=(l+r)/2;    if(k<0||k==r)return tree[step].n;    if(mid<k)return find(l,mid,-1,2*step)+find(mid+1,r,k,2*step+1);    return find(l,mid,k,2*step);}int main (void){    int s[11111],i,j,k,l,n,sum,Min;    while(scanf("%d",&n)!=EOF)    {        setit(0,n-1,1);        for(i=sum=0;i<n;i++)        scanf("%d",&s[i]),sum+=i-find(0,n-1,s[i],1),insert(0,n-1,s[i],1);        Min=sum;        for(i=1;i<n;i++)        {            sum+=n-1-2*s[i-1];            if(sum<Min)Min=sum;        }        printf("%d\n",Min);    }    return 0;}