首页 > ACM题库 > HDU-杭电 > HDU 1394 Minimum Inversion Number-线段树-[解题报告] C++
2013
12-09

HDU 1394 Minimum Inversion Number-线段树-[解题报告] C++

Minimum Inversion Number

问题描述 :

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.

输入:

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.

输出:

For each case, output the minimum inversion number on a single line.

样例输入:

10
1 3 6 9 0 8 5 7 4 2

样例输出:

16

#include<stdio.h>
 #define maxn 5010
 #define lson l,m,rt << 1
 #define rson m+1,r,rt << 1 | 1
 int tree[maxn<<2],x[maxn];
 int min(int a,int b)
 {
     return a<b?a:b;
 }
 void PushUP(int rt)
 {
     tree[rt]=tree[rt<<1]+tree[rt<<1|1];
 }
 void build(int l,int r,int rt)
 {
     tree[rt]=0;
     if(l==r)
         return ;
     int m=(l+r)>>1;
     build(lson);
     build(rson);
 }
 int query(int L,int R,int l,int r,int rt)
 {
     if(L<=l&&R>=r)
         return tree[rt];
     int m=(l+r)>>1;
     int ret=0;
     if(L<=m)
         ret+=query(L,R,lson);
     if(R>m)
         ret+=query(L,R,rson);
     return ret;
 }
 void update(int p,int l,int r,int rt)
 {
     if(l==r)
     {
         tree[rt]++;
         return ;
     }
     int m=(l+r)>>1;
     if(p<=m)
         update(p,lson);
     else
         update(p,rson);
     PushUP(rt);
 }
 int main()
 {
     int i,n,sum;
     while(scanf("%d",&n)!=EOF)
     {
         build(0,n-1,1);
         sum=0;
         for(i=0;i<n;i++)
         {
             scanf("%d",&x[i]);
             sum+=query(x[i],n-1,0,n-1,1);
             update(x[i],0,n-1,1);
         }
         int ret=sum;
         for(i=0;i<n;i++)
         {
             sum+=n-x[i]-1-x[i];
             ret=min(ret,sum);
         }
         printf("%d\n",ret);
     }
     return 0;
 }

解题报告转自:http://www.cnblogs.com/frog112111/archive/2012/09/01/2666970.html


  1. 第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。第2题,TCP不支持多播,多播和广播仅应用于UDP。所以B选项是不对的。