HDU 1394 線段樹or 樹狀數組~

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　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　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 個區間，裡麵包含有0~N-1的整數，允許你每次將序列的頭調到序列尾，讓你求出最小的逆序和，思路：先用線段樹or樹狀數組預先處理出原先的逆序和，然後再一個一個推出，每次操作後所產生的的新的逆序和，就拿範例來說，一開始的逆序和是22，如果將1調到末尾那裡的話，原先在1後面比1小的數有1個，這個數是0，而當1調到後面的時候，在1前面的比一大的數有8個，為3,6,9,8,5,7,4,2，那麼新序列的逆序和為22 - 1 + 8 = 29。以此類推 AC代碼：線段樹的方法，借鑒了notonlysuccess大神的姿勢：

Memory: 320 KB

Time: 78 MS

#include <cstdio>#include <iostream>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const 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;}

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後來我有用樹狀數組做了一下，發覺樹狀數組一般都要比線段樹要快~

Memory: 332 KB

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;}

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HDU 1394 線段樹or 樹狀數組~