Dating with girls(2)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2229    Accepted Submission(s): 638

Problem Description

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

7

Source

HDU 2009-5 Programming Contest

分析：这道题貌似被我搁置了很久没有理，回想了一下原因好像是因为当时wa了n发还是过不了但是觉得自己的方法没什么问题。现在重新看这个题目感觉没有一点印象了。

总体来说这个题目就是在一般的宽搜上多了一点就是墙壁会周期性（k）地消失，所以可以在标记数组增加一维，来表示step%k，因为在同一点余数不同，影响也不同。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2579代码清单：

#include<map>
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MAX = 105 ;

struct edge{ int x,y,step; };
int xy[4][2]={{0,-1},{-1,0},{0,1},{1,0}};
int T,r,c,k,sx,sy,ex,ey;
char str[MAX][MAX];
int vis[MAX][MAX][15];

bool check(edge v){
    if(v.x>=0&&v.x<r&&v.y>=0&&v.y<c&&!vis[v.x][v.y][v.step%k]){
        if(str[v.x][v.y]!='#')
            return true;
        if(str[v.x][v.y]=='#'&&v.step%k==0)
            return true;
    }
    return false;
}

void bfs(){
    queue<edge>q;
    while(q.size()) q.pop();
    memset(vis,0,sizeof(vis));
    edge p,w;
    p.x=sx;p.y=sy;p.step=0;
    vis[p.x][p.y][p.step%k]=1;
    q.push(p);
    while(q.size()){
        p=q.front();q.pop();
        if(p.x==ex&&p.y==ey){
            printf("%d\n",p.step);
            return ;
        }
        for(int i=0;i<4;i++){
            w.x=p.x+xy[i][0];
            w.y=p.y+xy[i][1];
            w.step=p.step+1;
            if(check(w)){
                vis[w.x][w.y][w.step%k]=1;
                q.push(w);
            }
        }
    }
    printf("Please give me another chance!\n");
}

int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%d",&r,&c,&k);
        for(int i=0;i<r;i++){
            scanf("%s",str[i]);
            for(int j=0;j<c;j++){
                if(str[i][j]=='Y'){ sx=i;sy=j; }
                if(str[i][j]=='G'){ ex=i;ey=j; }
            }
        }
        bfs();
    }return 0;
}