GCJ 2009 Round2 A Crazy Rows

/*
  这题之前一度会错意了，没理解“只允许交换相邻的两行”，意味着只能整行交换
  以及，最后要保证，每行最后出现1的位置，不得大于该行的行数
  
  又及，这题一开始...没看输入数据，以为输入数据之间是有空格的，相当于一个一个数字输入，后来看完数据格式才发现，其实应该输入一串字符串，然后分离为一个个数字
*/

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAX_N = 45;
int N;
int M[MAX_N][MAX_N]; // 矩阵

int a[MAX_N]; // a[i]表示第i行最后出现1的位置: 1~n-1 
void solve()
{
	fill (a, a + N, -1);
	int res = 0;
//	cout << "call solve and res = " << res << endl;
	for (int i = 0; i < N; i++)
	{
		a[i] = -1; // 如果第i行不含1的话，就当作-1 
		for (int j = 0; j < N; j++)
		if (M[i][j] == 1) a[i] = j;
	}
	
	for (int i = 0; i < N; i++)
	{
		int pos = -1; // 要移动到第i行的行
		for (int j = i; j < N; j++)
		{
			if (a[j] <= i)
			{
				pos = j;
				break;
			} 	
		}
		
		for (int j = pos; j > i; j--)
		{
//			cout << "nwo j = " << j << " and pos = " << pos << endl;
			swap(a[j], a[j - 1]);
			res++;
	//		cout << "now res = " << res << endl;
		} 
	}
	cout << res << endl;
} 
int main()
{
	freopen("E:\A-large.txt", "r", stdin);
	freopen("E:\output2.txt", "w", stdout);
	int k;
	cin >> k;
	for (int kase = 1; kase <= k; kase++ ) 
	{
		cin >> N;
		for (int i = 0; i < N; i++)
		{
			char s[MAX_N];
			cin >> s;
			for (int j = 0; j < N; j++)
			M[i][j] = s[j] - '0';
		}
//		for (int i = 0; i < N; i++)
//		for (int j = 0; j < N; j++)
//		cin >> M[i][j];
		
		printf("Case #%d: ",kase);  
		solve();
	}
	fclose(stdin);
	fclose(stdout);
	
	return 0;
}