有趣的逆序数——HDU1394

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11466    Accepted Submission(s): 7049

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

题意：就是给出n个数，如1，3，6，9，0，8，5，7，4，2这n=10个数，把第一个数放到最後面，其他不变，求该数列的逆序数的个数；之后又把第一个数放到最後面，。。。。

这样进行n-1次，求逆序数最小的序列的逆序数有多少个？

思路：只需要求出一个序列的逆序数，其他的可以根据sum=sum-a[i]+n-a[i]-1   得出；i为1~n-1,a[1]表示第二个数；求逆序数这里采用边输入边计算的方法，就比如这个例子：

输入1，就看2~9的范围已经输入了多少个数，这里是0个

输入3，就看4~9的范围已经输入了多少个数，这里是0个

.。。。

输入0，就看1~9的范围已经输入了多少个数，这里有1，3，6，9这4个

很明显，用线段数或树状数组能轻松解决；

还有就是归并排序能更快地解决！

数状数组：

#include<cstdio>#include<cstdlib>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>#include<string>#include<iostream>const int MAXN=5000+10;using namespace std;int c[MAXN];int a[MAXN];int lowbit(int x){return x&(-x);}void updata(int x, int num){while(x<=MAXN){c[x]+=num;x+=lowbit(x);}}int Sum(int x){int res=0;while(x>0){res+=c[x];x-=lowbit(x);}return res;}int main(){    //freopen("in.txt","r",stdin);    int n;    while(scanf("%d", &n)==1)    {        memset(c,0,sizeof(c));        int i,j;        int sum=0;        for(i=1; i<=n; i++){            scanf("%d", &a[i]);            j=a[i]+1;sum+=Sum(n+1)-Sum(j);0            updata(j,1);        }        int ans=sum;        for(i=1; i<n; i++){            sum=sum-a[i]+n-a[i]-1;            if(sum<ans) ans=sum;        }        printf("%d\n", ans);    }    return 0;}

线段数：

#include<cstdio>#include<cstdlib>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>#include<string>#include<iostream>const int MAXN=5000+10;using namespace std;int tree[MAXN*4];int sum;int a[MAXN];void update(int left, int right, int rt, int val){    if(left==right){        tree[rt]=1;        return;    }    int mid=(left+right)>>1;    if(val<=mid) update(left, mid, rt<<1, val);    else update(mid+1, right, rt<<1|1, val);    tree[rt]=tree[rt<<1]+tree[rt<<1|1];}void query(int left, int right, int rt, int l, int r){    if(l<=left && right<=r){        sum+=tree[rt];        return;    }    int mid=(left+right)>>1;    if(r<=mid) query(left, mid, rt<<1, l, r);    else if(l>mid) query(mid+1, right, rt<<1|1, l, r);    else{        query(left, mid, rt<<1, l, r);            query(mid+1, right, rt<<1|1, l, r);    }}int main(){    //freopen("in.txt","r",stdin);    int n;    while(scanf("%d", &n)==1)    {        memset(tree,0,sizeof(tree));        int i,j;        sum=0;        for(i=1; i<=n; i++){            scanf("%d", &a[i]);            j=a[i]+1;            query(1,n,1,j,n);                update(1,n,1,j);        }        int ans=sum;        for(i=1; i<n; i++){            sum=sum-a[i]+n-a[i]-1;            if(sum<ans) ans=sum;        }        printf("%d\n", ans);    }    return 0;}

归并排序：

#include<cstdio>#include<cstdlib>#include<cmath>#include<map>#include<queue>#include<stack>#include<vector>#include<algorithm>#include<cstring>#include<string>#include<iostream>const int MAXN=5000+10;using namespace std;int a[MAXN];int A[MAXN];int t[MAXN];int sum;void merge_sort(int *A,int x,int y,int *t){    if(y-x>1){        int m=x+(y-x)/2;        int p=x,q=m,i=x;        merge_sort(A,x,m,t);        merge_sort(A,m,y,t);        while(p<m || q<y){            if(q>=y || (p<m && A[p]<=A[q]))                t[i++]=A[p++];            else{                t[i++]=A[q++];                sum+=m-p;            }        }        for(i=x;i<y;i++){            A[i]=t[i];        }    }}int main(){    //freopen("in.txt","r",stdin);    int n;    while(scanf("%d", &n)==1)    {        int i,j;        sum=0;        for(i=1; i<=n; i++){            scanf("%d", &a[i]);        }        memcpy(A,a,sizeof(a));        merge_sort(a,1,n+1,t);        int ans=sum;        for(i=1; i<n; i++){            sum=sum-A[i]+n-A[i]-1;            if(sum<ans) ans=sum;        }        printf("%d\n", ans);    }    return 0;}