Minimum Inversion Number

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on 

HDU. Original ID: 

64-bit integer IO format: %I64d      Java class name: Main

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

101 3 6 9 0 8 5 7 4 2

Sample Output

16

Source

 

解题：求逆序数，很有意思的是算出原序列的逆序值后，只要加上 n - 1 - x - x就表示数列循环左移一位的逆序数值！只有从0开始且是连续的才成立。

 

1 #include 
        
          2 using namespace std; 3 const int maxn = 5005; 4  5 int d[maxn],tree[maxn<<2],n; 6 void update(int L,int R,int id,int v) { 7     if(id <= L && id >= R) { 8         tree[v]++; 9         return;10     }11     int mid = (L + R)>>1;12     if(id <= mid) update(L,mid,id,v<<1);13     if(id > mid) update(mid+1,R,id,v<<1|1);14     tree[v] = tree[v<<1] + tree[v<<1|1];15 }16 int query(int L,int R,int lt,int rt,int v){17     if(lt <= L && rt >= R) return tree[v];18     int mid = (L + R)>>1,ans = 0;19     if(lt <= mid) ans += query(L,mid,lt,rt,v<<1);20     if(rt > mid) ans += query(mid+1,R,lt,rt,v<<1|1);21     return ans;22 }23 int main() {24     while(~scanf("%d",&n)) {25         memset(tree,0,sizeof tree);26         int ans = 0,tmp = 0;27         for(int i = 0; i < n; ++i) {28             scanf("%d",d+i);29             tmp += query(0,n-1,d[i],n-1,1);30             update(0,n-1,d[i],1);31         }32         ans = tmp;33         for(int i = 0; i < n; ++i){34             tmp += n - 1 - d[i] - d[i];35             ans = min(ans,tmp);36         }37         printf("%d\n",ans);38     }39     return 0;40 }