HDU 2216 Game III(BFS)

Game III

[align=left]Problem Description[/align]
Zjt and Sara will take part in a game, named Game III. Zjt and Sara will be in a maze, and Zjt must find Sara. There are some strang rules in this maze. If Zjt move a step, Sara will move a step in opposite direction.
Now give you the map , you shold find out the minimum steps, Zjt have to move. We say Zjt meet Sara, if they are in the same position or they are adjacent .
Zjt can only move to a empty position int four diraction (up, left, right, down). At the same time, Sara will move to a position in opposite direction, if there is empty. Otherwise , she will not move to any position.
The map is a N*M two-dimensional array. The position Zjt stays now is marked Z, and the position, where Sara stays, is marked E.

> . : empty position
> X: the wall
> Z: the position Zjt now stay
> S: the position Sara now stay

Your task is to find out the minimum steps they meet each other.

[align=left]Input[/align]
The input contains several test cases. Each test case starts with a line contains three number N ,M (2<= N <= 20, 2 <= M <= 20 ) indicate the size of the map. Then N lines follows, each line contains M character. A Z and a S will be in the map as the discription above.

[align=left]Output[/align]
For each test case, you should print the minimum steps. “Bad Luck!” will be print, if they can't meet each other.

[align=left]Sample Input[/align]

4 4

XXXX

.Z..

.XS.

XXXX

4 4

XXXX

.Z..

.X.S

XXXX

4 4

XXXX

.ZX.

.XS.

XXXX

[align=left]Sample Output[/align]

1

1

Bad Luck!

[align=left]Answer[/align]
题解：这个BFS很有意思，跟典型的题目不同，它的目标点在动。当Z在移动的时候，S会往相反方向移动（如果能动）。就是这点，导致WA了两次，vis数组只开了二维来记录Z有没有走过，然后看了题解是要开四维数组，保存Z、S。然后第三组样例又一只跑不出来，发现要把判断是否访问语句放到判断S是否移动的后面才行。

#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#define PI acos(-1.0)
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
using namespace std;
//#define LOCAL
int n,m;
char mp[22][22];
bool vis[22][22][22][22];
int dir[4][2]={{1,0},{-1,0},{0,-1},{0,1}};
struct Node
{
int zx,zy;
int sx,sy;
int step;
}t,nn;
int bfs()
{
queue<Node> q;
while(!q.empty())q.pop();
vis[t.zx][t.zy][t.sx][t.sy]=1;
t.step=0;
q.push(t);
while(!q.empty())
{
t=q.front(),q.pop();
if(t.sx==t.zx&&abs(t.sy-t.zy)==1)return t.step;
if(t.sy==t.zy&&abs(t.sx-t.zx)==1)return t.step;
if(t.sx==t.zx&&t.sy==t.zy)return t.step;
for(int i=0;i<4;i++)
{
nn.zx=t.zx+dir[i][0];
nn.zy=t.zy+dir[i][1];
if(mp[nn.zx][nn.zy]=='X')continue;
if(nn.zx<0||nn.zx>=n||nn.zy<0||nn.zy>=m)continue;
nn.sx=t.sx-dir[i][0];
nn.sy=t.sy-dir[i][1];
if(nn.sx<0||nn.sx>=n||nn.sy<0||nn.sy>=m)nn.sx=t.sx,nn.sy=t.sy;
if(mp[nn.sx][nn.sy]=='X')nn.sx=t.sx,nn.sy=t.sy;
nn.step=t.step+1;
if(vis[nn.zx][nn.zy][nn.sx][nn.sy])continue;
vis[nn.zx][nn.zy][nn.sx][nn.sy]=1;
q.push(nn);
}
}
return -1;
}
int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
ios::sync_with_stdio(false);
while(cin>>n>>m)
{
msv;
for(int i=0;i<n;i++)cin>>mp[i];
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(mp[i][j]=='Z')t.zx=i,t.zy=j;
else if(mp[i][j]=='S')t.sx=i,t.sy=j;
}
int ans=bfs();
if(ans==-1)printf("Bad Luck!\n");
else printf("%d\n",ans);
}
return 0;
}

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