Minimum Inversion Number　　

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 題目大意就是給你 N 個(gè)區(qū)間，里面包含有0~N-1的整數(shù)，允許你每次將序列的頭調(diào)到序列尾，讓你求出最小的逆序和， 思路：先用線段樹(shù)or樹(shù)狀數(shù)組預(yù)處理出原先的逆序和，然后再一個(gè)一個(gè)推出，每次操作后所產(chǎn)生的的新的逆序和，就拿樣例來(lái)說(shuō)，一開(kāi)始的逆序和是22，如果將1調(diào)到末尾那里的話，原先在1后面比1小的數(shù)有1個(gè)，這個(gè)數(shù)是0，而當(dāng)1調(diào)到后面的時(shí)候，在1前面的比一大的數(shù)有8個(gè)，為3,6,9,8,5,7,4,2，那么新序列的逆序和為22 - 1 + 8 = 29。以此類推 AC代碼：線段樹(shù)的方法，借鑒了notonlysuccess大神的姿勢(shì)：

Memory:?320 KB

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Time:?78 MS

#include <cstdio> #include <iostream> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int maxn = 5555; int sum[maxn<<2]; void PushUp(int rt){sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void build(int l,int r,int rt){sum[rt] = 0;if(l == r) return ;int m = (l + r) >> 1;build(lson);build(rson);return ; } void update(int p,int l,int r,int rt){if(l == r){sum[rt]++;return ;}int m = (l + r)>>1;if(p <= m) update(p,lson);else update(p,rson);PushUp(rt); } int query(int L,int R,int l,int r,int rt){if(L <= l&&r <= R){return sum[rt];}int m = (l + r)>>1;int ret = 0;if(L <= m) ret += query(L,R,lson);if(R > m) ret += query(L,R,rson);return ret; } int x[maxn]; int main(){int n;while(~scanf("%d",&n)){build(0,n - 1,1);int sum = 0;for(int i = 0;i < n;i++){scanf("%d",&x[i]);sum += query(x[i],n - 1,0,n - 1,1);update(x[i],0,n - 1,1);}int res = sum;for(int i = 0;i < n;i++){sum += n - x[i] - x[i] - 1;res = min(res,sum);}printf("%d\n",res);}return 0; } View Code

后來(lái)我有用樹(shù)狀數(shù)組做了一下，發(fā)覺(jué)樹(shù)狀數(shù)組一般都要比線段樹(shù)要快~

Memory:?332 KB

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Time:?31 MS

#include <cstdio> #include <iostream> #include <cstring> using namespace std; const int maxn = 5555; int c[maxn]; int a[maxn]; int n; inline int lowbit(int x){return x&(-x); } int sum(int x){int ret = 0;while(x > 0){ret += c[x];x -= lowbit(x);}return ret; } void add(int x,int d){while(x <= n){c[x] += d;x += lowbit(x);} } int main(){while(~scanf("%d",&n)){memset(c,0,sizeof(c));memset(a,0,sizeof(a));int ans = 0;for(int i = 1;i <= n;i++){scanf("%d",&a[i]);ans += i - 1 - sum(a[i]);add(a[i] + 1,1);}//cout<<ans<<endl;int MIN = ans;for(int i = 1;i <= n;i++){MIN += n - (a[i] + 1) - (a[i]);ans = min(MIN,ans);}printf("%d\n",ans);}return 0; } View Code

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轉(zhuǎn)載于:https://www.cnblogs.com/jusonalien/p/4004973.html

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