toj 2815 Searching Problem
Time Limit: 1.0 Seconds Memory Limit: 65536K
Total Runs: 421 Accepted Runs: 177
It is said that the predecessors had hidden a lot of expensive treasures just before the death of their city hundreds of years ago. People's searching for it has never stopped. The only problem is that they can't tell the exact position of the treasure. What's more, there exists great dangerousness in searching. So although the treasure has been their place for countries, it's still safe.
You are one of the people who want to make a fortune. However, you're a very lucky man. You happen to get a secret map, and you know the map can lead you to the treasure directly.
The map is a maze. There are five symbols on the map. They are 'S', 'D', '$', '1' and '0'. 'S' is the point you enter into the maze. '$' stand for the treasure in the maze. '0' and 'D' are the points you can walk along, while '1' is walls that you can't go through. You shall start at the 'S' point, and pick up the treasure in the maze along the '0' and 'D' points, then come out along the same way.
It seems easy, and you'll become a wealthy man. But there is still a problem you have to solve. On every point marked as a 'D' there is a dog to protect the treasure. It's absurd that the dogs can live for countries without food! Just forget about it. So you decide to bring a gun with you. But how many bullets it will take? It's up to you to calculate here. Assume that you're a good shooter, and need only one bullet to shoot one dog.
Input
The first line of each test case is two integer R and C (1 ≤ R, C ≤ 50). R is the row number of the maze, and C is the column number of the maze. The following R lines will contain C characters for each. The R * C characters includes exactly one 'S' and one '$', the remains are '1', '0' and 'D'. It guarantees that the 'S' point is always on the edge of the maze.Two zeros indicate the end of the input.
Output
For each case, just print a line with the number K, which is the least number of bullets you need.Sample Input
4 5 00$00 00100 D00D0 S0001 4 5 00$00 01100 D00D0 S0001 4 5 DDDDD DD$DD D0D0D SDD0D 0 0
Sample Output
0 1 2
Problem Setter: Venice
Source: TJU Programming Contest 2007
#include <queue>
#define MAX 54
using namespace std;
int mark[MAX][MAX],a[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
char map[MAX][MAX];
typedef struct node
{
int x;
int y;
int step;
}Point;
queue<Point>Q;
Point start,p,temp;
int r,c,MM;
void Init()
{
int i,j,k;
for(i=0;i<r;i++)
for(j=0;j<c;j++)
{
cin>>map[i][j];
if(map[i][j]=='S')
{
start.x=i;
start.y=j;
start.step=0;
}
mark[i][j]=-1;
}
}
bool Bound(int x,int y)
{
if(x>=0&&y>=0&&x<r&&y<c)
return true;
else
return false;
}
void BFS()
{
int k,tx,ty;
while(!Q.empty())
Q.pop();
Q.push(start);
mark[start.x][start.y]=0;
MM=-1;
while(!Q.empty())
{
p=Q.front();
Q.pop();
if(map[p.x][p.y]=='$')
{
if(p.step<MM||MM==-1)
{
MM=p.step;
}
}
for(k=0;k<4;k++)
{
tx=p.x+a[k][0];
ty=p.y+a[k][1];
if(!Bound(tx,ty)||map[tx][ty]=='1')
continue;
if(map[tx][ty]=='0'||map[tx][ty]=='$')
{
if(mark[tx][ty]==-1||p.step<mark[tx][ty])
{
temp.x=tx;
temp.y=ty;
temp.step=p.step;
mark[tx][ty]=p.step;
Q.push(temp);
}
continue;
}
if(map[tx][ty]=='D')
{
temp.step=p.step+1;
if(mark[tx][ty]==-1||temp.step<mark[tx][ty])
{
temp.x=tx;
temp.y=ty;
//temp.step=p.step;
mark[tx][ty]=temp.step;
Q.push(temp);
}
continue;
}
}
}
printf("%d\n",MM);
}
int main()
{
while(scanf("%d%d",&r,&c)!=EOF)
{
if(r==0&&c==0)
break;
Init();
BFS();
}
return 0;
}
