hdu6318 离散化+树状数组求逆序数

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原文链接：http://www.cnblogs.com/the-way-of-cas/p/9380748.html

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#java

本文介绍了一道算法题目，通过使用树状数组优化逆序数的计算过程，旨在找到将序列变为有序所需的最小花费。文章详细解释了如何计算逆序数及优化策略。

Swaps and Inversions

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1609 Accepted Submission(s): 591

Problem Description

Long long ago, there was an integer sequence a.
Tonyfang think this sequence is messy, so he will count the number of inversions in this sequence. Because he is angry, you will have to pay x yuan for every inversion in the sequence.
You don't want to pay too much, so you can try to play some tricks before he sees this sequence. You can pay y yuan to swap any two adjacent elements.
What is the minimum amount of money you need to spend?
The definition of inversion in this problem is pair

Input

There are multiple test cases, please read till the end of input file.
For each test, in the first line, three integers, n,x,y, n represents the length of the sequence.
In the second line, n integers separated by spaces, representing the orginal sequence a.

Output

For every test case, a single integer representing minimum money to pay.

Sample Input

3 233 666

1 2 3

3 1 666

3 2 1

Sample Output

0 3

题意：花费x或者y交换序列中的任意相邻的一对数，使序列符合要求（本题从小到大）

思路：每次交换相邻的一组数，可以消除一个逆序数对，所以求出序列的逆序数，让后乘min(x,y);

树状数组：sum[i]=arr[i]+往前lowbit(i)个sum的和. lowbit(x)=x&-x. 表示末尾0的个数(二进制)

代码：

#include <stdio.h>    
#include <iostream>    
#include <queue>
#include <vector>
#include <algorithm>
#include <string.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define ll long long
#define maxx 100009
using namespace std;

int n, a[maxx], c[maxx];
int x, y;

struct node
{
    int v;
    int order;
}in[maxx];

int lowbit(int x) {
    return x & (-x);
}

void updata(int x, int val) {
    for (int i = x; i <= n; i += lowbit(i)) {
        c[i] += val;
    }
}
int getsum(int end) {
    int sum = 0;
    for (int i = end; i > 0; i -= lowbit(i))
    {
        sum += c[i];
    }
    return sum;
}

bool cmp(const node &a, const node &b) {
    if (a.v != b.v) return a.v < b.v;
    return a.order < b.order;
}//这个地方必须排order

int main() {
    fio;
    while (cin>>n>>x>>y)
    {
        memset(c, 0, sizeof c);
        memset(a, 0, sizeof a);
        memset(in, 0, sizeof in);

        for (int i = 1; i <= n; i++)
        {
            cin >> in[i].v;
            in[i].order = i;
        }

        sort(in + 1, in + n + 1, cmp);
        for (int i = 1; i <= n; i++) {
            a[in[i].order] = i;
        }

        ll sum = 0,ans;
        for (int i = 1; i <= n; i++) {
            updata(a[i], 1);
            sum += i - getsum(a[i]);
        }
        ans = sum * min(x, y);
        cout << ans << endl;
    }
    //getchar();
    return 0;
}