HDOJ_ACM_A计划

Problem Description

可怜的公主在一次次被魔王掳走一次次被骑士们救回来之后，而今，不幸的她再一次面临生命的考验。魔王已经发出消息说将在T时刻吃掉公主，因为他听信谣言说 吃公主的肉也能长生不老。年迈的国王正是心急如焚，告招天下勇士来拯救公主。不过公主早已习以为常，她深信智勇的骑士LJ肯定能将她救出。
现据密 探所报，公主被关在一个两层的迷宫里，迷宫的入口是S（0，0，0），公主的位置用P表示，时空传输机用#表示，墙用*表示，平地用.表示。骑士们一进入 时空传输机就会被转到另一层的相对位置，但如果被转到的位置是墙的话，那骑士们就会被撞死。骑士们在一层中只能前后左右移动，每移动一格花1时刻。层间的 移动只能通过时空传输机，且不需要任何时间。

 

Input

输入的第一行C表示共有C个测试数据，每个测试数据的前一行有三个整数N，M，T。 N，M迷宫的大小N*M（1 <= N,M <=10)。T如上所意。接下去的前N*M表示迷宫的第一层的布置情况，后N*M表示迷宫第二层的布置情况。

 

Output

如果骑士们能够在T时刻能找到公主就输出“YES”，否则输出“NO”。

 

Sample Input

1
5 5 14
S*#*.
.#...
.....
****.
...#.

..*.P
#.*..
***..
...*.
*.#..

 

Sample Output

YES

 

Idea

For this question, using flags[][], which mark wheather u passed or not, using dfs find the shortest path. U should care about more details, such as using flag to mark whether u find it, using outOfTime to mark whether the time is used up.

 

Code

#include <stdio.h>
#include <stdlib.h>
#include <queue>
using namespace std;
#define N 10
#define M 10
struct Position
{
    int x;
    int y;
    int step;
    int floornum;
};
struct
{
    char map[N + 2][M + 2];
    int flags[N + 2][M + 2];
}floors[2];
int dx[5] = {1, -1, 0, 0};
int dy[5] = {0, 0, 1, -1};
int main()
{
    int c, n, m, t, flag, outOfTime, floornum, result, i, j;
    Position source, des, pre, cur;
    scanf("%d", &c);
    while (c--)
    {
        queue<Position> q;
        scanf("%d %d %d", &n, &m, &t);
        getchar();
        for (floornum = 0; floornum < 2; floornum++)
        {
            for (i = 0; i < n; i++)
            {
                //input
                scanf("%s", floors[floornum].map[i]);
                //initialize
                for (j = 0; j < m; j++)
                    floors[floornum].flags[i][j] = 0;
            }
        }
        //printf("input successfully\n");
        source.x = 0;
        source.y = 0;
        source.step = 0;
        source.floornum = 0;
        q.push(source);
        flag = 0;        //mark whether find it or not
        outOfTime = 0;   //mark whether the time is used up or not
        floors[0].map[0][0] = 1;
        //printf("go to queue\n");
        while (!q.empty() && !flag && !outOfTime)
        {
            pre = q.front();
            q.pop();
            //printf("pop(): x = %d, y = %d, floornum = %d, step = %d\n", pre.x, pre.y, pre.floornum, pre.step);
            for (i = 0; i < 4; i++)
            {
                //get the current
                cur.x = pre.x + dx[i];
                cur.y = pre.y + dy[i];
                cur.step = pre.step + 1;
                cur.floornum = pre.floornum;
                //out of index
                if (cur.x < 0 || cur.x >= n || cur.y < 0 || cur.y >= m)
                    continue;
                //if the position has passed
                if (floors[cur.floornum].flags[cur.x][cur.y] == 1)
                    continue;
                //out of time
                if (cur.step > t)
                {
                    outOfTime = 1;
                    break;
                }
                //go the other floor
                if (floors[cur.floornum].map[cur.x][cur.y] == '#')
                {
                    floors[cur.floornum].flags[cur.x][cur.y] = 1;
                    cur.floornum = 1 - cur.floornum;
                }
                if (floors[cur.floornum].map[cur.x][cur.y] == 'P')
                {
                    flag = 1;
                    result = cur.step;
                    break;
                }
                //
                else if (floors[cur.floornum].map[cur.x][cur.y] == '.')
                {
                    floors[cur.floornum].flags[cur.x][cur.y] = 1;
                    q.push(cur);
                    //printf("push(): x = %d, y = %d, floornum = %d, step = %d\n", cur.x, cur.y, cur.floornum, cur.step);
                }
            }
        } 
        if (flag == 0)
            printf("NO\n");
        else if (outOfTime == 1)
            printf("NO\n");
        else if (flag == 1 && outOfTime == 0)
            printf("YES\n");
    }
    system("pause");
    return 0;
}