HDU 1394 Minimum Inversion Number（线段树+逆序数）
HDU 1394 Minimum Inversion Number（线段树+逆序数）
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
听过学长的讲课之后，还是不懂，后来发现自己把题意理解错了，orz。果然是被dp洗脑了，以为是求最大逆序数……研究了下别人的代码，发现其实很简单，和之前做的题差不多……代码如下~
#includeid*2+1,l,r);
return ans;
}
}//参照了线段树区间和的模板
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
build(1,0,n-1);
int ans=0;
int a[5005];
for(int i=0;i"%d",&a[i]);
ans+=query(1,a[i]+1,n-1);//每次输入以后求一次该数的逆序数，最终和即为所有逆序数的和
update(1,a[i]);//更新树
}
int Min=ans;
for(int i=0;i1-a[i]);//删除第i个数，会减少a[i]个逆序数，因为只有n-1个数，所以增加n-1-a[i]个逆序数；
Min=min(Min,ans);
}
printf("%d\n",Min);
}
return 0;
}