ZOJ - 1649 Rescue

题目：

Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison. 

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards. 

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.) 

 

Input

First line contains two integers stand for N and M. 

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend. 
//x是 guard，r不一定只有1个，#是墙
Process to the end of the file. 

 

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life." 

 

Sample Input

7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........

 

Sample Output

13

这个题目在OJ里面的掉了一些内容，我已经补上了。

广度优先搜索代码：

#include
#include
#include
using namespace std;

int n, m;
int  list[200][200];
int list2[200][200];		//list2只用来判断是不是敌人x
queue<int>q;

void f(int a, int b,int c,int d)
{
	if (a < 0 || a >= n)return;
	if (b < 0 || b >= m)return;
	if (list[a][b] == -1)
	{
		list[a][b] = list[c][d] + 1;
		q.push(a * 201 + b);
	}
	else if (list[a][b] == -3)
	{
		list[a][b] = list[c][d] + 2;
		q.push(a * 201 + b);
	}
	else if (list[a][b] > list[c][d] + 1)//因为没有用优先队列，所以有些格子需要重新进入队列
	{
		list[a][b] = list[c][d] + 1;
		if (list2[a][b])list[a][b]++;
		q.push(a * 201 + b);
	}
}

int main()
{	
	int a, b;
	char c;
	int x, y;
	while (cin>>n>>m)
	{
		while (q.size())q.pop();
		memset(list2, 0, sizeof(list2));
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < m; j++)
			{
				cin >> c;
				if (c == '.')list[i][j] = -1;
				else if (c == '#')list[i][j] = -2;
				else if (c == 'r')
				{
					list[i][j] = 0;
					q.push(i * 201 + j);
				}
				else if (c == 'x')
				{
					list[i][j] = -3;
					list2[i][j] = 1;
				}
				else
				{
					x = i;
					y = j;
					list[x][y] = -1;
				}
			}
		}
	while (!q.empty())//即使list[x][y]>0也不能结束循环，必须队列为空才能得到list[x][y]的最小值
	{
		a = q.front() / 201;
		b = q.front() % 201;
		q.pop();
		f(a + 1, b, a, b);
		f(a - 1, b, a, b);
		f(a, b + 1, a, b);
		f(a, b - 1, a, b);
	}
		if (list[x][y]>0)cout << list[x][y];
		else cout << "Poor ANGEL has to stay in the prison all his life.";
		cout << endl;
	}
	return 0;
}