HDU1394 Minimum Inversion Number
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5950 Accepted Submission(s): 3625
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
Author
CHEN, Gaoli
Source
Ignatius.L
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=5010; #define L(rt) (rt<<1) #define R(rt) (rt<<1|1) struct Tree{ int l,r; int sum; }tree[N<<2]; void PushUp(int rt){ tree[rt].sum=tree[L(rt)].sum+tree[R(rt)].sum; } void build(int L,int R,int rt){ tree[rt].l=L; tree[rt].r=R; if(tree[rt].l==tree[rt].r){ tree[rt].sum=0; return ; } int mid=(L+R)>>1; build(L,mid,L(rt)); build(mid+1,R,R(rt)); PushUp(rt); } void update(int id,int rt){ if(tree[rt].l==tree[rt].r){ tree[rt].sum++; return ; } int mid=(tree[rt].l+tree[rt].r)>>1; if(id<=mid) update(id,L(rt)); else if(id>=mid+1) update(id,R(rt)); PushUp(rt); } int query(int L,int R,int rt){ if(L<=tree[rt].l && tree[rt].r<=R){ return tree[rt].sum; } int mid=(tree[rt].l+tree[rt].r)>>1; int ans=0; if(R<=mid) ans+=query(L,R,L(rt)); else if(L>=mid+1) ans+=query(L,R,R(rt)); else{ ans+=query(L,mid,L(rt)); ans+=query(mid+1,R,R(rt)); } return ans; } int num[N]; int main(){ //freopen("input.txt","r",stdin); int n; while(~scanf("%d",&n)){ build(0,n-1,1); int tmp=0; for(int i=0;i<n;i++){ scanf("%d",&num[i]); tmp+=query(num[i],n-1,1); update(num[i],1); } int res=tmp; for(int i=0;i<n;i++){ tmp+=(n-1-num[i])-num[i]; res=min(res,tmp); } printf("%d\n",res); } return 0; }