HDU 1394 Minimum Inversion Number

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1394

题意：在序列 { 2, 4, 3, 1 } 中，逆序依次为 (2,1), (4,3), (4,1), (3,1)，因此该序列的逆序数为 4，题目给定一个长度为n的数组，每次可以把数组的前m个元素移动到数列的后面，所以共有n种数列，求这n种序列中，逆序数最小的一个

思路：（1）从第一种情况，递推剩下的情况，如果把a[1]放到末尾，则逆序数减少a[1]（从0到a[1]-1），逆序数增加n-(a[1]+1)（从a[1]+1到n）

（2）用sum数组标记到目前为止，这个数是否存，用线段树每次查询比它大的数有几个，初始的情况推出来了，剩下的情况和上面所用的方法一样

AC代码1：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>typedef long long ll;using namespace std;int a[5010];int main(){    int i,j;    int n;    while(scanf("%d",&n) != EOF)    {        for(i=0; i<n; i++)        {            scanf("%d",&a[i]);        }        int sum = 0;        for(i=0; i<n; i++)        {            for(j=i+1; j<n; j++)            {                if(a[i] > a[j])                {                    sum++;                }            }        }        int min1 = sum;        for(i=0; i<n; i++)        {            sum = sum - a[i] + n - 1 - a[i];            if(sum < min1)                min1 = sum;        }        printf("%d\n",min1);    }}

AC代码2：

#inlcude <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>typedef long long ll;using namespace std;int a[5010];int cf[5010];//也可以每次在merge1函数里面每次开一个数组，这样比较省内存int lol[5010];int sum;void merge1(int b,int e){    int i = b;    int mid = (b + e) / 2;    int j = mid + 1;    int temp = b;    while(i <= mid && j <= e)    {        if(a[i] <= a[j])        {            cf[temp++] = a[i];            i++;        }        else        {            cf[temp++] = a[j];            j++;            sum += mid - i + 1;//i到mid都比这个数大        }    }    while(i <= mid)    {        cf[temp++] = a[i];        i++;    }    while(j <= e)    {        cf[temp++] = a[j];        j++;    }    for(i=b; i<=e; i++)    {        a[i] = cf[i];    }}void merge_sort(int b,int e){    if(b < e)    {        int mid = (b + e) / 2;        merge_sort(b,mid);        merge_sort(mid+1,e);        merge1(b,e);    }}int main(){    int i;    int n;    while(scanf("%d",&n) != EOF)    {        sum = 0;        for(i=0; i<n; i++)        {            scanf("%d",&a[i]);            lol[i] = a[i];        }        merge_sort(0,n-1);        int min1 = sum;        for(i=0; i<n; i++)        {            sum = sum - lol[i] + n - 1 - lol[i];            if(sum < min1)                min1 = sum;        }        printf("%d\n",min1);    }}

AC代码3：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long LL;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;const int maxn = 5010;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(L <= m) ans += query(L,R,lson);    if(R > m) ans += query(L,R,rson);    return ans;}void updata(int p,int l,int r,int rt){    if(l == r)    {        sum[rt]++;        return;    }    int m = (l + r) >> 1;    if(p <= m) updata(p,lson);    if(p > m) updata(p,rson);    pushup(rt);}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",&a[i]);            sum += query(a[i]+1,n-1,0,n-1,1);            updata(a[i],0,n-1,1);        }        int min1 = sum;        for(int i=0; i<n; i++)        {            sum = sum - a[i] + n - 1 - a[i];            if(sum < min1)                min1 = sum;        }        printf("%d\n",min1);    }    return 0;}