Minimum Inversion Number 【线段数】
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
来源: http://acm.hdu.edu.cn/showproblem.php?pid=1394
大意:给出一个从0到n-1的序列,逐次把首数字移到尾部,问,最小逆序数?
题解:
这里的逆序数和线性代数中的逆序数是一个概念,某个数前面出现比其大的数,称为逆序,总序列中逆序的个数,称为逆序数。
用线段树求第一次输入的序列的逆序数,然后用公式来遍历所有转移情况。
#include<stdio.h> #include<string.h> #include<string.h> #include<algorithm> #define M 5001 using namespace std; struct Node{ int a; int b; int sum; }t[3*M]; int p[M],total; /* 创建范围为x~y的线段树 */ void make(int x,int y,int n){ t[n].a = x; t[n].b = y; t[n].sum = 0; if(x != y){ int mid = (x + y)/2; make(x,mid,2*n); make(mid+1,y,2*n+1); } } /* 返回 x~y 区间内的个数sum */ int query(int x,int y,int n){ if(x <= t[n].a && y >= t[n].b){ return t[n].sum; }else{ int mid = (t[n].a + t[n].b)/2; if(x > mid){ query(x,y,2*n+1); }else if( y <= mid){ query(x,y,2*n); }else{ return query(x,y,2*n)+query(x,y,2*n+1); } } } /* 从最高级区间开始往下面的具有x的区间的sum */ void update(int x,int n){ t[n].sum++; if(t[n].a == x && t[n].b == x){ return; } int mid = (t[n].a + t[n].b)/2; if(x > mid){ update(x,2*n+1); }else{ update(x,2*n); } } int main(){ int n,m,a[M]; while(~scanf("%d",&n)){ make(0,n-1,1); int total = 0; for(int i=0;i<n;i++){ scanf("%d",&a[i]); total += query(a[i],n-1,1); update(a[i],1); } int ans = total; for(int i=0;i<n;i++){ total = total - a[i] + (n - a[i] - 1); ans = min(ans,total); } printf("%d\n",ans); } return 0; } /* total = total - a[i] + (n - a[i] - 1); 比如1 3 6 9 0 8 5 7 4 2 比1小的数有0个,后面比1大的数有8个,1放到后面,少了1个逆序数,又多了8个逆序数 比3小的数有3个,后面比3大的数有6个,3放到后面,少了3个逆序数,又多了6个逆序数 。。。 */