POJ 1804 Brainman

本题求将一个序列排序成有序序列所需要的最少交换次数，交换只能相邻交换

可以知道，最少次数就等于原序列的逆序对数

首先

交换相邻的数，最多只能把逆序对数减一，
而排序好的序列中逆序对数为0
未排序好的序列中逆序对数不为0
换句话说，至少需要交换“逆序对数”次才能把序列排序

其次，只要序列不是有序的序列，
就必然存在 i 使得 a[i]>a[i+1]
这样，只要交换 i 和 i+1
就能使逆序对数减一，
换句话说，存在交换“逆序对数”次就能把序列排序的算法

上界=下界，所以答案就是逆序对数...

Brainman

Time Limit:1000MS

Memory Limit:30000K

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

/* Author yan
 * POJ 1804
 * Brainman
*/
#include<stdio.h>
int value[1000];
int cnt;

int main()
{
	int test;
	int n;
	int i,j,k;
	//freopen("input","r",stdin);
	scanf("%d",&test);
	for(i=1;i<=test;i++)
	{
		cnt=0;
		scanf("%d",&n);
		for(j=0;j<n;j++) scanf("%d",&value[j]);
		for(j=0;j<n;j++)
			for(k=j+1;k<n;k++)
				if(value[j]>value[k]) cnt++;
		printf("Scenario #%d:/n%d/n/n",i,cnt);
	}
	return 0;
}