逆序数2 HDOJ 1394 Minimum Inversion Number

题目传送门

/*
    求逆序数的四种方法
*/

/*
    1. O(n^2) 暴力+递推 法:如果求出第一种情况的逆序列，其他的可以通过递推来搞出来，一开始是t[1],t[2],t[3]....t[N]
    它的逆序列个数是N个，如果把t[1]放到t[N]后面，逆序列个数会减少t[1]个，相应会增加N-(t[1]+1)个  
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAX_N = 10000 + 10;
const int INF = 0x3f3f3f3f;
int a[MAX_N];
int num[MAX_N];

int main(void)      //HDOJ 1394 Minimum Inversion Number
{
    //freopen ("inC.txt", "r", stdin);
    int n;

    while (~scanf ("%d", &n))
    {
        memset (num, 0, sizeof (num));
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &a[i]);
            //a[n+i] = a[i];
        }
        int t = 0;  int sum = 0;
        for (int i=1; i<=n; ++i)    //先求解最初的数列逆序数
        {
            for (int j=i+1; j<=n; ++j)
            {
                if (a[i] > a[j])
                {
                    sum++;
                }
            }
        }
        //printf ("%d
", sum);
        int ans = INF;
        for (int i=1; i<=n; ++i)        //更新sum，找最小
        {
            sum = sum - a[i] + (n - a[i] - 1);      //the next line contains a permutation of the n integers from 0 to n-1. 
            //printf ("%d
", res);                 //从0到n-1的整数  所以这里用a[i]  读题不仔细 。。。。
            ans = min (sum, ans);
        }
        printf ("%d
", ans);
    }

    return 0;
}

/*
    2. 归并算法
*/
#include <cstdio>
#include <algorithm>
using namespace std;

const int MAX_N = 10000 + 10;
const int INF = 0x3f3f3f3f;
int a[MAX_N];
int b[MAX_N];
int L[MAX_N/2+10], R[MAX_N/2+10];
int cnt = 0;

void Merge(int *a, int p, int q, int r)
{
    int n1 = q - p + 1;
    int n2 = r - q;
    int i, j;

    for (i=1; i<=n1; ++i)
    {
        L[i] = a[p+i-1];
    }
    for (j=1; j<=n2; ++j)
    {
        R[j] = a[q+j];
    }
    L[n1+1] = R[n2+1] = INF;
    i = j = 1;
    for (int k=p; k<=r; ++k)
    {
        if (L[i] <= R[j])
        {
            a[k] = L[i++];
        }
        else
        {
            a[k] = R[j++];
            cnt += n1 - i + 1;
        }
    }
}

void MergeSort(int *a, int p, int r)
{
    if (p < r)
    {
        int q = (p + r) / 2;
        MergeSort (a, p, q);
        MergeSort (a, q+1, r);
        Merge (a, p, q, r);
    }
}

int main(void)      //HDOJ 1394 Minimum Inversion Number
{
    //freopen ("inC.txt", "r", stdin);
    int n;
    while (~scanf ("%d", &n) && n)
    {
        for (int i=1; i<=n; ++i)
        {
            scanf ("%d", &a[i]);
            b[i] = a[i];
        }
        MergeSort (a, 1, n);
        //printf ("%d
", cnt);
        int ans = INF;
        for (int i=1; i<=n; ++i)        //注意a[i]已排序
        {
            cnt = cnt - b[i] + (n - b[i] - 1);
            ans = min (cnt, ans);
        }
        printf ("%d
", ans);
        cnt = 0;
    }

    return 0;
}

/*
    3. nlogn 线段树-单点更新：更新比a[i]大的个数
*/
#include <cstdio>
#include <algorithm>
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1 | 1

const int MAX_N = 5000 + 10;
const int INF = 0x3f3f3f3f;
int a[MAX_N];
int sum[MAX_N << 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);
}

void update(int p, int l, int r, int rt)
{
    if (l == r)
    {
        sum[rt]++;      //记录次数
        return ;
    }
    int m = (l + r) >> 1;
    if (p <= m)
    {
        update (p, lson);
    }
    else
        update(p, rson);
    pushup (rt);
}

int query(int ql, int qr, int l, int r, int rt)
{
    if (ql <= l && r <= qr)
    {
        return sum[rt];
    }
    int m = (l + r) >> 1;
    int ans = 0;
    if (ql <= m)    ans += query (ql, qr, lson);
    if (qr > m)     ans += query (ql, qr, rson);

    return ans;
}

int main(void)      //HDOJ 1394 Minimum Inversion Number
{
    //freopen ("inC.txt", "r", stdin);
    int n;
    while (~scanf ("%d", &n))
    {
        //memset (num, 0, sizeof (num));
        build (0, n-1, 1);
        int sum = 0;
        for (int i=1; i<=n; ++i)   
        {
            scanf ("%d", &a[i]);
            sum += query (a[i], n-1, 0, n-1, 1);
            update (a[i], 0, n-1, 1);
        }
        int ans = sum;
        for (int i=1; i<=n; ++i)        
        {
            sum = sum - a[i] + (n - a[i] - 1);            
            ans = std::min (sum, ans);
        }
        printf ("%d
", ans);
    }

    return 0;
}

/*
    4. 树状数组
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAXN=5050;
int c[MAXN];
int a[MAXN];
int n;

int lowbit(int x)
{
    return x&(-x);
}

void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}

int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}

int main()      //HDOJ 1394 Minimum Inversion Number
{
    //freopen ("inC.txt", "r", stdin);
    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            ans+=sum(n)-sum(a[i]);
            add(a[i],1);
        }
        int Min=ans;
        for(int i=1;i<=n;i++)
        {
            ans+=n-a[i]-(a[i]-1);
            if(ans<Min)Min=ans;
        }
        printf("%d
",Min);
    }
    
    return 0;
}