Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5892 Accepted Submission(s): 3587
Problem 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.
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 using namespace std; 7 #define lson l, m, rt<<1 8 #define rson m+1, r, rt<<1|1 9 const int maxn = 5555; 10 int sum[maxn<<2]; 11 void PushUP(int rt){ 12 sum[rt] = sum[rt<<1] + sum[rt<<1|1]; 13 } 14 void build(int l, int r, int rt){ 15 if (l == r) {sum[rt] = 0; return;} 16 int m= (l + r) >> 1; build(lson); build(rson); 17 PushUP(rt); 18 } 19 void update(int p, int l, int r, int rt){ 20 if (l == r) {sum[rt]++; return;} 21 int m = (l + r) >> 1; 22 if (p <= m) update(p, lson); else update(p, rson); 23 PushUP(rt); 24 } 25 int query(int L, int R, int l, int r, int rt){ 26 if (L <= l && R >= r) {return sum[rt];} 27 int m = (l + r) >> 1, ret = 0; 28 if (L <= m) ret += query(L, R, lson); 29 if (R > m) ret += query(L, R, rson); 30 return ret; 31 } 32 int x[maxn]; 33 int main(void){ 34 #ifndef ONLINE_JUDGE 35 freopen("hdu1394.in", "r", stdin); 36 #endif 37 int n; 38 while (~scanf("%d", &n)){ 39 int sum = 0; build(0, n-1, 1); 40 for (int i = 0; i < n; ++i){ 41 scanf("%d", x+i); sum += query(x[i],n-1,0,n-1,1); 42 update(x[i], 0, n-1, 1); 43 } 44 int ret = sum; 45 for (int i = 0; i < n; ++i){ 46 sum += n - 1 - x[i] - x[i]; 47 ret = min(ret, sum); 48 } 49 printf("%d\n", ret); 50 } 51 return 0; 52 }
学习树状数组做的题目，还是想了一下午，自己模拟一下就知道怎么回事了，其实也不难。至于怎么建树，怎么更新树状数组，自己画一个树状的图就可以理解了。就是先用线段树求出原有序列的逆序数，然后再求出其他排列的逆序数，这里有一个技巧，就是45行到48行的代码，自己体会一下就Ok了，很巧妙的。
