poj 1804 （nyoj 117）Brainman ：归并排序求逆序数

Brainman

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 7810		Accepted: 4261

Description

Background

Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just by glancing at them. And he can even count Poker cards. Charlie would love to be able to do cool things like that, too. He
wants to beat his brother in a similar task.

Problem

Here's what Charlie thinks of. Imagine you get a sequence of N numbers. The goal is to move the numbers around so that at the end the sequence is ordered. The only operation allowed is to swap two adjacent numbers. Let us try an example:

Start with: 2 8 0 3

swap (2 8) 8 2 0 3

swap (2 0) 8 0 2 3

swap (2 3) 8 0 3 2

swap (8 0) 0 8 3 2

swap (8 3) 0 3 8 2

swap (8 2) 0 3 2 8

swap (3 2) 0 2 3 8

swap (3 8) 0 2 8 3

swap (8 3) 0 2 3 8

So the sequence (2 8 0 3) can be sorted with nine swaps of adjacent numbers. However, it is even possible to sort it with three such swaps:

Start with: 2 8 0 3

swap (8 0) 2 0 8 3

swap (2 0) 0 2 8 3

swap (8 3) 0 2 3 8

The question is: What is the minimum number of swaps of adjacent numbers to sort a given sequence?Since Charlie does not have Raymond's mental capabilities, he decides to cheat. Here is where you come into play. He asks you to write a computer program for him
that answers the question. Rest assured he will pay a very good prize for it.

Input

The first line contains the number of scenarios.

For every scenario, you are given a line containing first the length N (1 <= N <= 1000) of the sequence,followed by the N elements of the sequence (each element is an integer in [-1000000, 1000000]). All numbers in this line are separated by single blanks.

Output

Start the output for every scenario with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the minimal number of swaps of adjacent numbers that are necessary to sort the given sequence.
Terminate the output for the scenario with a blank line.

Sample Input

4

4 2 8 0 3

10 0 1 2 3 4 5 6 7 8 9

6 -42 23 6 28 -100 65537

5 0 0 0 0 0

Sample Output

Scenario #1:

3

Scenario #2:

0

Scenario #3:

5

Scenario #4:

0

归并排序求出逆序数，拿的以前写好的模板，速度还不错，不过发现之前模板有一个错误，就是合并完以后并没有释放new的空间，在POJ上运行没问题，就是内存大一点，但是在NYOJ上，如果不delete就会MLE，两题的格式不一样，以下是POJ的AC代码，NYOJ需要修改输出格式才能AC

#include<stdio.h>

#include<iostream>

using namespace std;

int array[1000001];

long long flag = 0;

void merg(int head, int tail)

{

	int mid = (tail + head) / 2 + 1;

	int * new_array = new int[(tail - head) + 1];

	int top1 = head;

	int top2 = mid;

	int i;

	for(i = 0; top1 < mid  && top2 <= tail ; i++)

	{

		if(array[top1] > array[top2])

		{

			new_array[i] = array[top2];

			top2 ++;

		}

		else

		{

			new_array[i] = array[top1];

			flag += top2 - (mid);

			top1 ++;

		}

	}

	if(top1 == mid && top2 <= tail)

	{

		while(top2 <= tail)

			new_array[i++] = array[top2++];

	}

	else if(top1 != mid && top2 > tail)

	{

		while(top1 < mid)

		{

			new_array[i++] = array[top1++];

			flag += tail - (mid) + 1;

		}

	}

	memcpy(&array[head], new_array, sizeof(int) * (tail - head + 1) );

	delete new_array;

}

void mergsort(int head, int tail)

{

	if(head >= tail)

		return ;

	mergsort(head, (head + tail) / 2);

	mergsort((head + tail) / 2 + 1, tail);

	merg(head, tail);

}

int main()

{

	int n;

	int m;

//	freopen("test.txt", "r", stdin);

	scanf("%d", &m);

	int j;

	for(j = 1; j <= m; j++)

	{

		printf("Scenario #%d:\n", j);

		scanf("%d", &n);

		int i;

		flag = 0;

		for(i = 0; i < n; i++)

			scanf("%d", &array[i]);

		mergsort(0, n - 1);

		printf("%lld\n\n", flag);

	}

	return 0;

}

巴特西

poj 1804 （nyoj 117）Brainman ：归并排序求逆序数

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巴特西

poj 1804 （nyoj 117）Brainman ： 归并排序求逆序数

最新文章

热门文章

poj 1804 （nyoj 117）Brainman ：归并排序求逆序数