HDU 1394 Minimum Inversion Number【线段树，归并排序，树状数组】

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 15476 Accepted Submission(s): 9441

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, output the minimum inversion number on a single line.

[align=left]Sample Input[/align]

10
1 3 6 9 0 8 5 7 4 2

[align=left]Sample Output[/align]

16

给出一个序列，然后每次都把最后边的数放在最前面，由此形成多个序列，求给定的一个数列形成的所有的新数列的最小逆序对数

逆序不难求得，主要是求出所有的逆序，当然进行n次线段树的操作是不可取的，所以要特殊考虑规律.....

假如 某个序列的逆序数为 s ，这个序列的最后一个元素是 am，一共n 个数，那么，在这个数的前面 有 n-am个数比am大，有 am-1个数比am 小，那么进过操作之后，逆序数会减少n-am，增加 am-1，可得，新得到的数列的逆序数为 s+n-2*am-1.....

此题用归并求逆序比较好用，因为这个题的数据比较特殊，用线段树还是比较方便的，另外还有树状数组.......

93MS

#include<stdio.h>
#include<string.h>
int sum[4000005],x[1000005];
int find(int rt,int l,int r,int a,int b)
{
if(l>=a&&r<=b)
{
return sum[rt];
}
int mid=(l+r)>>1,tp=rt<<1,res=0;
if(mid>=a)
{
res+=find(tp,l,mid,a,b);
}
if(mid<b)
{
res+=find(tp|1,mid+1,r,a,b);
}
return res;
}
void update(int rt,int l,int r,int p)
{
if(l==r)
{
++sum[rt];
return;
}
int mid=(l+r)>>1,tp=rt<<1;
if(p<=mid)
{
update(tp,l,mid,p);
}
else
{
update(tp|1,mid+1,r,p);
}
sum[rt]=sum[tp]+sum[tp|1];
}
int main()
{
int n;
while(~scanf("%d",&n))
{
int cnt=0;
memset(sum,0,sizeof(sum));
for(int i=0;i<n;++i)
{
scanf("%d",&x[i]);
cnt+=find(1,0,n-1,x[i],n-1);
update(1,0,n-1,x[i]);
}
int tp=cnt;
for(int i=0;i<n;++i)
{
tp+=(n-2*x[i]-1);
if(tp<cnt)
{
cnt=tp;
}
}
printf("%d\n",cnt);
}
return 0;
}

归并排序：（46MS）

#include<stdio.h>
#include<string.h>
int cnt,x[5005],t[5005];
void merge(int l,int r)
{
int mid=(l+r)>>1;
int a=l,b=mid+1,c=l;
while(a<=mid&&b<=r)
{
if(x[a]<=x[b])
{
t[c++]=x[a++];
}
else
{
cnt+=mid-a+1;
t[c++]=x[b++];
}
}
while(a<=mid)
{
t[c++]=x[a++];
}
while(b<=r)
{
t[c++]=x[b++];
}
for(int i=l;i<=r;++i)
{
x[i]=t[i];
}
}

void sort(int l,int r)
{
if(l<r)
{
int mid=(l+r)>>1;
sort(l,mid);sort(mid+1,r);
merge(l,r);
}
}

int main()
{
int n,y[5005];
//freopen("shuju.txt","r",stdin);
while(~scanf("%d",&n))
{
//memset(sum,0,sizeof(sum));
for(int i=0;i<n;++i)
{
scanf("%d",&x[i]);
y[i]=x[i];
}
cnt=0;
sort(0,n-1);
int tp=cnt;
for(int i=0;i<n;++i)
{
tp+=(n-2*y[i]-1);
if(tp<cnt)
{
cnt=tp;
}
}
printf("%d\n",cnt);
}
return 0;
}

用树状数组的方法有空再补上......