HDU 1394 线段树or 树状数组~

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　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

1 3 6 9 0 8 5 7 4 2

Sample Output

题目大意就是给你 N 个区间，里面包含有0~N-1的整数，允许你每次将序列的头调到序列尾，让你求出最小的逆序和，

思路：先用线段树or树状数组预处理出原先的逆序和，然后再一个一个推出，每次操作后所产生的的新的逆序和，就拿样例来说，一开始的逆序和是22，如果将1调到末尾那里的话，原先在1后面比1小的数有1个，这个数是0，而当1调到后面的时候，在1前面的比一大的数有8个，为3,6,9,8,5,7,4,2，那么新序列的逆序和为22 - 1 + 8 = 29。以此类推

AC代码：线段树的方法，借鉴了notonlysuccess大神的姿势：

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|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
",res);
    }
    return 0;
}

View Code

后来我有用树状数组做了一下，发觉树状数组一般都要比线段树要快~

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
",ans);
    }
    return 0;
}

View Code

额继续努力吧骚年~~~