BFS(最短路) HDOJ 4308 Saving Princess claire_

题目传送门

题意：一个(r*c<=5000)的迷宫，起点'Y‘，终点'C'，陷阱‘#’，可行路‘*’(每走一个，*cost)，传送门P，问Y到C的最短路

分析：一道最短路问题，加了传送门的功能，那么第一次走到P时是最短的路径，之后再到P都不会比第一次短，所以将所有P看成一个点，走过就vis掉，剩下就是简单的BFS了

收获：1、复习BFS求最短路　　2. memset不能乱用，尤其是数组很大时，容易爆内存

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015-8-22 11:56:55
* File Name     :I.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
typedef pair<int, int> P;
const int N = 5e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int dx[4] = {-1, 1, 0, 0};
int dy[4] = {0, 0, -1, 1};
struct Node	{
    int x, y, step;
};
Node t[N/10];
char maze[N][N];
bool vis[N][N];
int n, m, cost, tot;
int sx, sy, ex, ey;

bool judge(int x, int y)	{
    if (x < 1 || x > n || y < 1 || y > m || vis[x][y] || maze[x][y] == '#')	return false;
    return true;
}

int BFS(void)	{
//	memset (vis, false, sizeof (vis));
//  memset (d, INF, sizeof (d));
	for (int i=1; i<=n; ++i)	{
		for (int j=1; j<=m; ++j)	vis[i][j] = false;
	}
    vis[sx][sy] = true;
    queue<Node> Q;	Q.push (Node {sx, sy, 0});

    while (!Q.empty ())	{
		Node u = Q.front ();	Q.pop ();
		for (int i=0; i<4; ++i)	{
			int tx = u.x + dx[i], ty = u.y + dy[i];
			if (!judge (tx, ty))	continue;
			vis[tx][ty] = true;
			if (maze[tx][ty] == 'C')	{
				return u.step;
			}
			if (maze[tx][ty] == 'P')	{
				for (int i=1; i<=tot; ++i)	{
					t[i].step = u.step;
					Q.push (t[i]);
					vis[t[i].x][t[i].y] = true;
				}
			}
			else	{
				Q.push (Node {tx, ty, u.step + 1});
			}
		}
    }

	return -1;
}

int main(void)    {
    while (scanf ("%d%d%d", &n, &m, &cost) == 3)	{
        sx = sy = 0;	ex = ey = 0;	tot = 0;
        for (int i=1; i<=n; ++i)	{
            scanf ("%s", maze[i] + 1);
            for (int j=1; j<=m; ++j)	{
                if (maze[i][j] == 'Y')	{
                    sx = i, sy = j;
                } else if (maze[i][j] == 'C')	{
                    ex = i, ey = j;
                } else if (maze[i][j] == 'P')	{
                    t[++tot].x = i;	t[tot].y = j;
                }
            }
        }

		int res = BFS ();
        if (res != -1)	{
            printf ("%d
", res * cost);
        } else	puts ("Damn teoy!");
    }

    return 0;
}