hdu1394线段树点修改，区间求和

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

 

一开始看这题时，弄好好久才把题目意思弄懂。

 

 

题目意思是求最小逆序数，而逆序数就是前面的数大于后面的数，而一串数字又可以移动，现在要求逆序数最小的是有多少。

 

我们可以先把每个数的当前位置有多少个位置求出来，如a1,a2,a3,a4......an分别对应有b1,b2,b3,b4......bn个逆序数。求出一开始的逆序数sn，但把第一个数放在最后时：

a2,a3,a4......an,a1  这时的逆序数就sn - a1 + n - a1 - 1 保留最小值即可，然后遍历整个数组就可以求出最小逆序数了。

下面贴下我写的 线段树+暴力 代码

 

线段树代码：

#include <iostream>
#include <cstdio>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
using namespace std;
const int maxn = 5e3+10;

int min(int a, int b){
    return a>b?b:a;
}
int sum[maxn<<2], a[maxn];
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);
}
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 ans = 0;
    if(m >= L) ans +=query(L, R, lson);
    if(m < R) ans += query(L, R, rson);
    return ans;
}
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 main(){
    int n;
    while(cin>>n){
        int sn = 0;
        build(0, n - 1, 1);
        for(int i = 0; i < n; i++){
            scanf("%d",&a[i]);
            sn += query(a[i], n - 1, 0, n -1, 1);
            update(a[i], 0, n - 1, 1);
        }
        int ans = sn;
        for(int i = 0; i < n; i++){
            sn = sn - a[i] + n - a[i] - 1;
            if(sn < ans){
                ans = sn;
            }
        }
        cout << ans << endl;
    }
    return 0;
}

暴力代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[5005],b[5005];
int main(){
    int n;
    while(cin>>n){
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for(int i = 0; i < n; i++){
            scanf("%d",&a[i]);
        }
        int sum = 0;
        for(int i = 0; i < n; i++){
            for(int j = i+1; j < n; j++){
                if(a[i] > a[j]){
                    b[i]++;
                }
            }
            sum += b[i];
        }
        int minx = sum;
        for(int i = 0; i < n; i++){
            sum = sum - a[i] + (n - a[i] - 1);
            if(sum < minx){
                minx = sum;
            }
        }
        cout << minx << endl;
    }
    return 0;
}