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[HDU] 1394 Minimum Inversion Number [线段树求逆序数]

2024-08-04 17:51:37   220人阅读

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11788    Accepted Submission(s): 7235


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.
 

 

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
 

 

Author
CHEN, Gaoli
 

 

Source
ZOJ Monthly, January 2003
 

 

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Ignatius.L
 
题解：这题的维护信息为每个数是否已经出现。每次输入后，都从该点的值到n-1进行查询，每次发现出现了一个数，由于是从该数的后面开始找的，这个数肯定是比该数大的。那就是一对逆序数，然后逆序数+1.最后求完所有的逆序数之后，剩下的就可以递推出来了。因为假如目前的第一个数是x，那当把他放到最后面的时候，少的逆序数是本来后面比他小的数的个数。多的逆序数就是放到后面后前面比他大的数的个数。因为所有数都是从0到n-1.所以比他小的数就是x，比他大的数就是n-1-x。这样的话每次的逆序数都可以用O(1)的时间计算出来。然后找最小的时候就可以了。
 
 1 #include<cstdio> 2 #include<algorithm> 3  4 using namespace std; 5  6 #define  lson  l , m , rt << 1 7 #define  rson  m + 1 , r , rt << 1 | 1 8  9 const int maxn = 5000 + 311;10 int sum[maxn<<2];11 12 13 void PushUp(int rt) 14 {15     sum[rt]=sum[rt<<1]+sum[rt<<1|1];16 }17 18 void build(int l,int r,int rt)19 {20     int m;21     22     sum[rt]=0;23     if(l==r) {24         return;25     }26     m=(l+r)>>1;27     build(lson);28     build(rson);29 }30 31 int query(int L,int R, int l, int r,int rt)32 {33     int m,ret;34     35     if(L<=l && r<=R) {36         return sum[rt];37     }38     39     m=(l+r) >> 1;40     ret=0; 41     if(L<=m) ret+=query(L,R,lson);42     if(R>m) ret+=query(L,R,rson);43     44     return ret;45 } 46 47 void Updata(int p,int l,int r,int rt)48 {49     int m;50     if (l==r) {51         sum[rt]++;52         return ;53     }54     m=(l+r)>>1;55     if(p<=m) Updata(p,lson);56     else Updata(p,rson);57     58     PushUp(rt); 59 }60 61 int main()62 {63     int n,sum,x[maxn];64     65     while(~scanf("%d",&n)) {66         sum=0;67         build(0,n-1,1);68         69         for(int i = 0;i < n;i++) {70             scanf("%d",&x[i]);71             sum+=query(x[i],n-1,0,n-1,1);72             Updata(x[i],0,n-1,1);73         }74         int ret=sum;75         for(int i=0;i<n;i++) {76             sum+=n-x[i]-x[i]-1;77             ret=min(ret,sum);78         }79         80         printf("%d\n",ret);81     }82     83     return 0;84 }

 

[HDU] 1394 Minimum Inversion Number [线段树求逆序数]


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