SGU Inversions（线段树）

最新推荐文章于 2015-06-24 09:46:00 发布

佩儿

最新推荐文章于 2015-06-24 09:46:00 发布

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本文链接：https://blog.youkuaiyun.com/u011606503/article/details/39505191

本文介绍了一种计算给定整数序列中逆序对数量的方法，包括输入格式、输出要求以及具体实现过程。

Inversions

Time Limit: 250MS

Memory Limit: 4096KB

64bit IO Format: %I64d & %I64u

Description

180. Inversions

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

input: standard
output: standard

There are N integers (1<=N<=65537) A1, A2,.. AN (0<=Ai<=10^9). You need to find amount of such pairs (i, j) that 1<=i<j<=N and A[i]>A[j].

Input

The first line of the input contains the number N. The second line contains N numbers A1...AN.

Output

Write amount of such pairs.

Sample test(s)

Input

5 2 3 1 5 4

Output

代码：

#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define MAXN 66537

struct node
{
    int x,y;
} a[MAXN];
long long sum[MAXN*4];
int b[MAXN];

bool cmp(const node aa,const node bb)
{
    return aa.x<bb.x;
}

void build(int s,int t,int start)
{
    sum[start]=0;
    if(s==t) return ;
    int mid=(s+t)/2;
    build(s,mid,start*2);
    build(mid+1,t,start*2+1);
}

void update(int p,int s,int t,int start)
{
    if(s==t)
    {
        sum[start]++;
        return ;
    }
    int mid=(s+t)/2;
    if(p<=mid) update(p,s,mid,start*2);
    else update(p,mid+1,t,start*2+1);
    sum[start]=sum[start*2]+sum[start*2+1];
}

long long query(int l,int r,int s,int t,int start)
{
    if(l<=s&&t<=r)
        return sum[start];
    int mid=(s+t)/2;
    long long ret=0;
    if(l<=mid) ret+=query(l,r,s,mid,start*2);
    if(r>mid) ret+=query(l,r,mid+1,t,start*2+1);
    return ret;
}

int main()
{
    int n;
    long long sum1;
    while(scanf("%d",&n)!=EOF)
    {
        build(1,n,1);
        sum1=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].x);
            a[i].y=i;
        }
        sort(a+1,a+n+1,cmp);
        int ans=1;
        b[a[1].y]=1;
        for(int i=2;i<=n;i++)
        {
             if(a[i-1].x==a[i].x) b[a[i].y]=ans;
             else
             {
                 ans++;
                 b[a[i].y]=ans;
             }
        }
        for(int i=1; i<=n; i++)
        {
            //cout<<b[i]<<' ';
            sum1+=query(b[i]+1,n,1,n,1);
            update(b[i],1,n,1);
        }
        printf("%I64d\n",sum1);
    }
    return 0;
}