Dating with girls(2)【HDU - 2579 】

最新推荐文章于 2019-05-11 16:40:14 发布

原创最新推荐文章于 2019-05-11 16:40:14 发布 · 291 阅读

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本文介绍了一个有趣的迷宫约会问题：玩家需要在特定时间内找到迷宫中的女孩，期间部分障碍物会周期性消失。通过广度优先搜索算法实现，考虑了障碍物随时间变化的影响。

If you have solved the problem Dating with girls(1).I think you can solve this problem too.This problem is also about dating with girls. Now you are in a maze and the girl you want to date with is also in the maze.If you can find the girl, then you can date with the girl.Else the girl will date with other boys. What a pity!
The Maze is very strange. There are many stones in the maze. The stone will disappear at time t if t is a multiple of k(2<= k <= 10), on the other time , stones will be still there.
There are only ‘.’ or ‘#’, ’Y’, ’G’ on the map of the maze. ’.’ indicates the blank which you can move on, ‘#’ indicates stones. ’Y’ indicates the your location. ‘G’ indicates the girl's location . There is only one ‘Y’ and one ‘G’. Every seconds you can move left, right, up or down.

Input

The first line contain an integer T. Then T cases followed. Each case begins with three integers r and c (1 <= r , c <= 100), and k(2 <=k <= 10).
The next r line is the map’s description.

Output

For each cases, if you can find the girl, output the least time in seconds, else output "Please give me another chance!".

Sample Input

1
6 6 2
...Y..
...#..
.#....
...#..
...#..
..#G#.

Sample Output

题解：

求从Y到G的最短时间，'.'是路，可以走，如果时间%k==0，’#‘可以走，否则不可以走。

需要用到广搜，并注意判断石头是否可以行走。

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
int n,m,k,sx,sy,ex,ey;
char mapp[120][120];
int dir[4][2]= {0,1,0,-1,1,0,-1,0},s[120][120][12];
struct node
{
    int x,y,step;
};

void bfs()
{
    node now,temp;
    now.x=sx;
    now.y=sy;
    now.step=0;
    queue<node>q;
    memset(s,0,sizeof(s));
    s[sx][sy][0]=1;
    q.push(now);
    while(!q.empty())
    {
        temp=q.front();
        q.pop();
        if(temp.x==ex&&temp.y==ey)
        {
            printf("%d\n",temp.step);
            return ;
        }
        for(int i=0; i<4; i++)
        {
            int xx,yy;
            xx=temp.x+dir[i][0];
            yy=temp.y+dir[i][1];
            if(xx>=0&&xx<n&&yy>=0&&yy<m&&(mapp[xx][yy]!='#'||(temp.step+1)%k==0)&&!s[xx][yy][(temp.step+1)%k])
            {
                now.step=temp.step+1;
                now.x=xx;
                now.y=yy;
                s[xx][yy][(temp.step+1)%k]=1;
                q.push(now);
            }
        }
    }
    printf("Please give me another chance!\n");
}
int main()
{
    int f;
    scanf("%d",&f);
    while(f--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=0; i<n; i++)
        {
            scanf("%s",mapp[i]);
            for(int j=0; j<m; j++)
            {
                if(mapp[i][j]=='Y')
                {
                    sx=i;
                    sy=j;
                }
                else if(mapp[i][j]=='G')
                {
                    ex=i;
                    ey=j;
                }
            }
        }
        bfs();
    }
    return 0;
}