hdu 1394 Minimum Inversion Number
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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.
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
101 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
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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 | 1const 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); }}
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