【hdu1394】Minimum Inversion Number——逆序对

题目：

C - Minimum Inversion Number Time Limit:1000MS Memory
Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status
Practice HDU 1394 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

描述：

给定一个0-n-1长度为n的数列（n <= 5000），求这个数列的所有排列中逆序对数最小的，所有排列是指每次将数列的第一个数移动到最后一个，重复n-1次。

题解：

首先，逆序对有三种解法，树状数组和线段树是常规解法，但是由于这次的数据是连续的而且从0开始，说明我们不需要对数据进行离散化，这样的话除了常规做法外，由于数据范围不大，可以根据逆序对的定义直接暴力求解。
接下来对于每次移动，将第一个移动到最后一个，意味着原逆序对数要加上所有比这个数大的数，减去所有比这个数小的数，枚举每一次的变化即可。

代码：

树状数组版

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5005;int c[maxn],a[maxn],n;int lowbit(int x){    return x & (-x);}void add(int x,int val){    for(;x <= n;x += lowbit(x))c[x] += val;}int cal(int x){    int ans = 0;    for(;x > 0;x -= lowbit(x))ans += c[x];    return ans;}int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%d",&n))    {        memset(c,0,sizeof(c));        for(int i = 1;i <= n;i++)            scanf("%d",&a[i]);        int sum = 0,ans;        for(int i = n;i > 0;i--)        {            sum += cal(a[i]);            add(a[i]+1, 1);        }        ans = sum;        for(int i = 1;i <= n;i++)        {            sum += n - 1 - a[i] * 2;            ans = min(ans, sum);        }        printf("%d\n",ans);    }    return 0;}

线段树版

#include <cstdio>#include <cstring>#include <algorithm>#define ls t<<1#define rs (t<<1)|1using namespace std;const int maxn = 5005;struct segmentTree{    int l,r,val;   }tree[maxn*3];int fa[maxn],a[maxn],n;void update(int t){    if(t == 1)return;    t >>= 1;    tree[t].val = tree[ls].val + tree[rs].val;    update(t);}void build(int l,int r,int t){    tree[t].l = l;    tree[t].r = r;    tree[t].val = 0;    if(l == r)    {        fa[l] = t;        return ;    }    build(l, (l+r)/2, ls);    build((l+r)/2+1, r, rs);}long long query(int l,int r,int t){    long long sum = 0;    if(tree[t].l >= l && tree[t].r <= r) return tree[t].val;    int mid = (tree[t].l + tree[t].r) / 2;    if(mid >= l) sum += query(l,r,ls);    if(mid < r) sum += query(l,r,rs);    return sum;}int main(){    //freopen("in.txt","r",stdin);    while(~scanf("%d",&n))    {        long long sum = 0,ans;        build(1,n,1);        for(int i = 0;i < n;i++)        {            scanf("%d",&a[i]);            sum += query(a[i]+1, n, 1);            tree[fa[a[i]+1]].val = 1;            update(fa[a[i]+1]);        }        ans = sum;        for(int i = 0;i < n;i++)        {            sum += n - 1 - a[i] * 2;            ans = min(ans,sum);        }        printf("%lld\n",ans);    }    return 0;}

暴力直接求版

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5005;int a[maxn];int main(){    //freopen("in.txt","r",stdin);    int n;    while(~scanf("%d",&n))    {        long long ans = 0,res = 0;        for(int i = 0;i < n;i++)scanf("%d",&a[i]);        for(int i = 0;i < n;i++)                  for(int j = i+1;j < n;j++)                if(a[i] > a[j]) ans++;        res = ans;        for(int i = 0;i < n;i++)        {            ans -= a[i]-(n-1-a[i]);            res = min(res,ans);        }        printf("%lld\n",res);             }    return 0;}