HDU1394 Minimum Inversion Number

　　原题传送：http://acm.hdu.edu.cn/showproblem.php?pid=1394

　　求逆序数。

　　题目上来一看，发现直接暴力可过，复杂度O(n²)，300+ms，代码简短得要命：

#include <stdio.h>

int a[5005];

inline int min(int x, int y)
{
    return x < y ? x : y;
}

int main()
{
    int i, j, n, ans, m;
    while(scanf("%d", &n) != EOF)
    {
        ans = 0;
        for(i = 1; i <= n; i ++)
            scanf("%d", &a[i]);
            
        for(i = 1; i <= n; i ++)
            for(j = i + 1; j <= n; j ++)
                if(a[i] > a[j])
                    ans ++;
        m = ans;
        for(i = 1; i < n; i ++)
            ans = min(ans, m = m + n - a[i] - 1 - a[i]);
        printf("%d\n", ans);
    }
    return 0;
}

　　其实这道题复杂度都耗在了第一次求总逆序数上了，后来想想，发现可以用nlog(n)的归并排序搞，15～30ms。

#include <stdio.h>
#define N 5005
int a[N], b[N], c[N];

inline int min(int x, int y)
{
    return x < y ? x : y;
}

int cal(int s, int t)
{
    int i, j, k, mid;
    int res;
    if(t - s == 1)
        return 0;
    mid = (s + t) >> 1;
    res = cal(s, mid) + cal(mid, t);
    i = s, j = mid, k = s;
    while(i < mid && j < t)
    {
        if(a[i] <= a[j])
            b[k ++] = a[i ++], res += j - mid;
        else
            b[k ++] = a[j ++];
    }
    while(i < mid) b[k ++] = a[i ++], res += j - mid;
    while(j < t) b[k ++] = a[j ++];
    for(i = 0; i < t - s; i ++)
        a[s + i]= b[s + i];
    return res;
}

int main()
{
    int i, n, ans, m;
    while(scanf("%d", &n) != EOF)
    {
        for(i = 1; i <= n; i ++)
        {
            scanf("%d", &a[i]);
            c[i] = a[i];
        }
        ans = cal(1, n + 1);
        m = ans;
        for(i = 1; i < n; i ++)
            ans = min(ans, m = m + n - c[i] - 1 - c[i]);
        printf("%d\n", ans);
    }
    return 0;
}

　　其实我是想用线段树写的，30ms。

#include <stdio.h>
#include <string.h>
#define lson (cur << 1)
#define rson (cur << 1 | 1)
#define N 5005

int tree[N << 2], a[N];

inline int min(int x, int y)
{
    return x < y ? x : y;
}

void pushup(int cur)
{
    tree[cur] = tree[lson] + tree[rson];
}

void query(int cur, int l, int r, int x, int y, int &ans)
{
    if(l >= x && r <= y)
    {
        ans += tree[cur];
        return ;
    }
    int mid = (l + r) >> 1;
    if(mid >= x)
        query(lson, l, mid, x, y, ans);
    if(mid + 1 <= y)
        query(rson, mid + 1, r, x, y, ans);
}

void update(int cur, int l, int r, int loc)
{
    if(l == r)
    {
        tree[cur] = 1;
        return ;
    }
    int mid = (l + r) >> 1;
    if(mid >= loc)
        update(lson, l, mid, loc);
    else
        update(rson, mid + 1, r, loc);
    pushup(cur);
}

int main()
{
    int n, i, v, ans, sum;
    while(scanf("%d", &n) != EOF)
    {
        memset(tree, 0, sizeof tree);
        for(sum = 0, i = 1; i <= n; i ++)
        {
            scanf("%d", &a[i]);
            v = 0;
            query(1, 1, n, a[i] + 1, n, v);
            sum += v;
            update(1, 1, n, a[i]);
        }
        ans = sum;
        for(i = 1; i < n; i ++)
            ans = min(ans, sum = sum + n - a[i] - 1 - a[i]);
        printf("%d\n", ans);
    }
    return 0;
}