hdu2807---The Shortest Path

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The Shortest Path

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2552 Accepted Submission(s): 807

Problem Description

There are N cities in the country. Each city is represent by a matrix size of M*M. If city A, B and C satisfy that A*B = C, we say that there is a road from A to C with distance 1 (but that does not means there is a road from C to A).
Now the king of the country wants to ask me some problems, in the format:
Is there is a road from city X to Y?
I have to answer the questions quickly, can you help me?

Input

Each test case contains a single integer N, M, indicating the number of cities in the country and the size of each city. The next following N blocks each block stands for a matrix size of M*M. Then a integer K means the number of questions the king will ask, the following K lines each contains two integers X, Y(1-based).The input is terminated by a set starting with N = M = 0. All integers are in the range [0, 80].

Output

For each test case, you should output one line for each question the king asked, if there is a road from city X to Y? Output the shortest distance from X to Y. If not, output "Sorry".

Sample Input

Sample Output

  
   1
Sorry

Source

HDU 2009-4 Programming Contest

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暴力模拟矩阵运算然后floyd

/*************************************************************************
    > File Name: hdu2807.cpp
    > Author: ALex
    > Mail: 405045132@qq.com 
    > Created Time: 2015年01月03日 星期六 12时13分36秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 85;
const int inf = 0x3f3f3f3f;

int dp[N][N];
int n, m;
struct martix
{
	int	mat[N][N];
	void clear()
	{
		memset (mat, 0, sizeof(mat));
	}
}city[N][N], node[N];

void floyd()
{
	for (int k = 1; k <= n; ++k)
	{
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= n; ++j)
			{
				dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]);
			}
		}
	}
}

void martix_mutiply(int a, int b)
{
	city[a][b].clear();
	for (int i = 1; i <= m; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			for (int k = 1; k <= m; ++k)
			{
				city[a][b].mat[i][j] += node[a].mat[i][k] * node[b].mat[k][j];
			}
		}
	}
}

bool is_same(int a, int b, int c)
{
	for (int i = 1; i <= m; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			if (city[a][b].mat[i][j] != node[c].mat[i][j])
			{
				return 0;
			}
		}
	}
	return 1;
}

int main()
{
	while (~scanf("%d%d", &n, &m))
	{
		if (!n && !m)
		{
			break;
		}
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= m; ++j)
			{
				for (int k = 1; k <= m; ++k)
				{
					scanf("%d", &node[i].mat[j][k]);
				}
			}
		}
		memset (dp, inf, sizeof(dp));
		for (int i = 1; i <= n; ++i)
		{
			dp[i][i] = 0;
		}
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= n; ++j)
			{
				if (i != j)
				{
					martix_mutiply(i, j);
				}
			}
		}
		for (int i = 1; i <= n; ++i)
		{
			for (int j = 1; j <= n; ++j)
			{
				if (i == j)
				{
					continue;
				}
				for (int k = 1; k <= n; ++k)
				{
					if (i == k || j == k)
					{
						continue;
					}
					if (is_same(i, j, k))
					{
						dp[i][k] = 1;
					}
				}
			}
		}
		floyd();
		int q;
		scanf("%d", &q);
		while (q--)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			if (dp[x][y] == inf)
			{
				printf("Sorry\n");
			}
			else
			{
				printf("%d\n", dp[x][y]);
			}
		}
	}
	return 0;
}