hdu 1394 Minimum Inversion Number-优快云博客

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11143 Accepted Submission(s): 6854

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.
逆序数是给定一个序列a1,a2...,an，这个序列中一对数 (ai, aj)满足i<j则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:
给定一个序列a1,a2...,an如果我们移动第一个 m >= 0个数字到序列的末尾，我们将得到另一个序列，那一共有n种序列
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.

你要去写一个程序在以上的队列中找到最小的逆序数

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.

输入包含多组数据。每组数据包含两行：第一行包括一个整数n (n <= 5000);下一行包括n个数的一种排列

Output

For each case, output the minimum inversion number on a single line.

对于每组数据，输出最小的逆序数

Sample Input

  
   10
1 3 6 9 0 8 5 7 4 2

Sample Output

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

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Ignatius.L | We have carefully selected several similar problems for you: 1540 1542 1255 1828 1541

学习大牛的线段树里面的题目。。。求逆序数，用线段树插入过程中就求出来第一组逆序数，其他组的用数学方法推。

#include <stdio.h>
#include <string.h>
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

const int maxn = 5555;
int sum[maxn << 2];
int x[maxn];

int min(int a, int b){
    return a < b ? a : b;
}

void PushUp(int rt){
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void build(int l, int r, int rt){
    int m;
    sum[rt] = 0;
    if (l == r)
        return;
    m = (l + r) >> 1;
    build(lson);
    build(rson);
    PushUp(rt);
}

int query(int L, int R, int l, int r, int rt){
    int ret = 0;
    int m;
    if (L <= l && r <= R){
        return sum[rt];
    }
    m = (l + r ) >> 1;
    if (L <= m)
        ret += query(L, R, lson);
    
    if (R > m)
        ret += query(L, R, rson);

    return ret;

}

void update(int p,  int l, int r, int rt){
    int m = 0;
    if (l == r){
        sum[rt] ++; 
        return;
    }    
    m = (l + r) >> 1;
    if (p <= m){
        update(p, lson);
    }else update(p, rson);
    PushUp(rt);
    return;
}
int main(void){

    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    int i;
    int sum_num = 0;
    int ret;
    int n;
    while(scanf("%d", &n) != EOF){
        build(0, n - 1, 1);
        sum_num = 0;
        for (i = 0;i < n;i ++){
            scanf("%d", &x[i]);
            sum_num += query(x[i], n - 1, 0, n - 1, 1);
            update(x[i], 0, n - 1, 1);
        }
        ret = sum_num;
        for (i = 0;i < n;i ++){
            sum_num += n - x[i] - x[i] - 1;
            ret = min(sum_num, ret);
        }
        printf("%d\n", ret);
    }

}