标签：des style http color os io java strong ar
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Minimum Inversion NumberTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11157 Accepted Submission(s): 6865
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
Sample Output
Author
CHEN, Gaoli
Source
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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=5000+10;
int n,ans,cnt;
int a[maxn];
void pre_gao()
{
cnt=0;
for(int i=0;i<n-1;i++)
for(int j=i+1;j<n;j++)
if(a[i]>a[j])
cnt++;
}
int main()
{
while(~scanf("%d",&n))
{
ans=INF;
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
pre_gao();
for(int i=0;i<n;i++)
{
cnt=cnt-a[i]+n-1-a[i];
ans=min(ans,cnt);
}
printf("%d\n",ans);
}
return 0;
}
/*
5
5 4 6 7 3
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=5000+10;
int tree[maxn<<2],n,a[maxn];
void push_up(int dex)
{
tree[dex]=tree[dex<<1]+tree[dex<<1|1];
}
void buildtree(int l,int r,int dex)
{
if(l==r)
{
tree[dex]=0;
return;
}
int mid=(l+r)/2;
buildtree(l,mid,dex<<1);
buildtree(mid+1,r,dex<<1|1);
push_up(dex);
}
int Query(int L,int R,int l,int r,int dex)
{
if(L<=l&&R>=r)
return tree[dex];
int mid=(l+r)/2;
if(R<=mid) return Query(L,R,l,mid,dex<<1);
else if(L>mid) return Query(L,R,mid+1,r,dex<<1|1);
else return Query(L,R,l,mid,dex<<1)+Query(L,R,mid+1,r,dex<<1|1);
}
void Update(int k,int l,int r,int dex)
{
if(l==r)
{
tree[dex]++;
return;
}
int mid=(l+r)/2;
if(k<=mid) Update(k,l,mid,dex<<1);
else Update(k,mid+1,r,dex<<1|1);
push_up(dex);
}
int main()
{
int cnt,ans;
while(~scanf("%d",&n))
{
cnt=0;
buildtree(1,n,1);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]=a[i]+1;
cnt+=Query(a[i],n,1,n,1);
Update(a[i],1,n,1);
}
ans=cnt;
for(int i=1;i<n;i++)
{
cnt=cnt-(a[i]-1)+n-a[i];
ans=min(ans,cnt);
}
printf("%d\n",ans);
}
return 0;
}
/*
5
2 3 0 1 4
*/
hdu1394 Minimum Inversion Number(线段树单点更新||暴力)
标签：des style http color os io java strong ar
原文地址：http://blog.csdn.net/u014303647/article/details/38975745
