POJ 1657 解题报告

最新推荐文章于 2018-11-13 19:36:25 发布

thestoryofsnow

最新推荐文章于 2018-11-13 19:36:25 发布

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分类专栏： POJ

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本文介绍了一种使用广度优先搜索（BFS）算法来解决棋盘上不同棋子从一个位置移动到另一个位置所需的最少步数的问题。具体考虑了国王、皇后、车和象四种棋子的移动方式。

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BFS简单题。

thestoryofsnow

1657

Accepted

156K

0MS

C++

3173B

/* 
ID: thestor1 
LANG: C++ 
TASK: poj1657 
*/
#include <iostream>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <limits>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <cassert>

using namespace std;

int kdirections[8][2] = {{1, -1}, {1, 0}, {1, 1}, {0, -1}, {0, 1}, {-1, -1}, {-1, 0}, {-1, 1}};
int rdirections[4][2] = {{1, 0}, {0, -1}, {0, 1}, {-1, 0}};
int bdirections[4][2] = {{1, -1}, {1, 1}, {-1, -1}, {-1, 1}};

bool isInBorder(int nr, int nc)
{
	return 0 <= nr && nr < 8 && 0 <= nc && nc < 8;
}

std::vector<pair<int, int> > neighbours(const int r, const int c, int type, const int steps[][8])
{
	std::vector<pair<int, int> > nbs;
	// king
	if (type == 0)
	{
		for (int i = 0; i < 8; ++i)
		{
			int nr = r + kdirections[i][0], nc = c + kdirections[i][1];
			if (isInBorder(nr, nc) && steps[nr][nc] < 0)
			{
				nbs.push_back(make_pair(nr, nc));
			}
		}
	}
	// queen
	else if (type == 1)
	{
		for (int i = 0; i < 8; ++i)
		{
			int nr = r + kdirections[i][0], nc = c + kdirections[i][1];
			while (isInBorder(nr, nc) && steps[nr][nc] < 0)
			{
				nbs.push_back(make_pair(nr, nc));
				nr += kdirections[i][0], nc += kdirections[i][1];
			}
		}
	}
	// rook
	else if (type == 2)
	{
		for (int i = 0; i < 4; ++i)
		{
			int nr = r + rdirections[i][0], nc = c + rdirections[i][1];
			while (isInBorder(nr, nc) && steps[nr][nc] < 0)
			{
				nbs.push_back(make_pair(nr, nc));
				nr += rdirections[i][0], nc += rdirections[i][1];
			}
		}
	}
	// Bishop
	else
	{
		assert(type == 3);
		for (int i = 0; i < 4; ++i)
		{
			int nr = r + bdirections[i][0], nc = c + bdirections[i][1];
			while (isInBorder(nr, nc) && steps[nr][nc] < 0)
			{
				nbs.push_back(make_pair(nr, nc));
				nr += bdirections[i][0], nc += bdirections[i][1];
			}
		}
	}

	return nbs;
}

int main()
{
	int T;	
	scanf("%d", &T);

	int steps[8][8];

	for (int t = 0; t < T; ++t)
	{
		char r1c, r2c;
		int r1, c1, r2, c2;
		scanf(" %c%d %c%d", &r1c, &c1, &r2c, &c2);
		r1 = r1c - 'a', r2 = r2c - 'a';
		c1--, c2--;

		// (r1, c1), (r2, c2)
		const pair<int, int> dest = make_pair(r2, c2);

		for (int type = 0; type < 4; ++type)
		{
			queue<pair<int, int> > que;
			que.push(make_pair(r1, c1));
			memset(steps, -1, 8 * 8 * sizeof(int));
			steps[r1][c1] = 0;

			while (!que.empty())
			{
				pair<int, int> pos = que.front();
				que.pop();

				if (pos == dest)
				{
					break;
				}

				int r = pos.first, c = pos.second;
				int step = steps[r][c];

				std::vector<pair<int, int> > nbs = neighbours(r, c, type, steps);
				for (int i = 0; i < nbs.size(); ++i)
				{
					int nr = nbs[i].first, nc = nbs[i].second;
					steps[nr][nc] = step + 1;
					que.push(nbs[i]);
				}
			}

			if (steps[r2][c2] >= 0)
			{
				printf("%d", steps[r2][c2]);
			}
			else
			{
				printf("Inf");
			}

			if (type != 3)
			{
				printf(" ");
			}
			else
			{
				printf("\n");
			}
		}

	}
	return 0;  
}