poj 3083 搜索 DFS,BFS
Children of the Candy Corn
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11225 | Accepted: 4850 |
Description
The cornfield maze is a popular Halloween treat. Visitors are shown the entrance and must wander through the maze facing zombies, chainsaw-wielding psychopaths, hippies, and other terrors on their quest to find the exit.
One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.
One popular maze-walking strategy guarantees that the visitor will eventually find the exit. Simply choose either the right or left wall, and follow it. Of course, there's no guarantee which strategy (left or right) will be better, and the path taken is seldom the most efficient. (It also doesn't work on mazes with exits that are not on the edge; those types of mazes are not represented in this problem.)
As the proprieter of a cornfield that is about to be converted into a maze, you'd like to have a computer program that can determine the left and right-hand paths along with the shortest path so that you can figure out which layout has the best chance of confounding visitors.
Input
Input to this problem will begin with a line containing a single integer n indicating the number of mazes. Each maze will consist of one line with a width, w, and height, h (3 <= w, h <= 40), followed by h lines of w characters each that represent the maze layout. Walls are represented by hash marks ('#'), empty space by periods ('.'), the start by an 'S' and the exit by an 'E'.
Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').
You may assume that the maze exit is always reachable from the start point.
Exactly one 'S' and one 'E' will be present in the maze, and they will always be located along one of the maze edges and never in a corner. The maze will be fully enclosed by walls ('#'), with the only openings being the 'S' and 'E'. The 'S' and 'E' will also be separated by at least one wall ('#').
You may assume that the maze exit is always reachable from the start point.
Output
For each maze in the input, output on a single line the number of (not necessarily unique) squares that a person would visit (including the 'S' and 'E') for (in order) the left, right, and shortest paths, separated by a single space each. Movement from one square to another is only allowed in the horizontal or vertical direction; movement along the diagonals is not allowed.
Sample Input
2 8 8 ######## #......# #.####.# #.####.# #.####.# #.####.# #...#..# #S#E#### 9 5 ######### #.#.#.#.# S.......E #.#.#.#.# #########
Sample Output
37 5 5 17 17 9
Source
South Central USA 2006
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <queue>
using namespace std;
const int maxn = 100+10;
int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};
int dl[][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int dr[][2] = {{0, 1}, {-1, 0}, {0, -1}, {1, 0}};
int sx, sy, ex, ey, n, m;
char G[maxn][maxn];
struct Pos {
int x, y, s;
};
int dfs(int x, int y, int d, int step, int dir[][2]) {
for(int i=0; i<4; i++) {
int j = ((d-1+4)%4+i)%4;
int nx = x+dir[j][0];
int ny = y+dir[j][1];
if(nx == ex && ny == ey) return step+1;
if(nx < 0 || ny < 0 || nx > n || ny > m) continue;
if(G[nx][ny] == '#') continue;
return dfs(nx, ny, j, step+1, dir);
}
}
int BFS(int sx, int sy) {
bool vis[maxn][maxn];
memset(vis, false, sizeof(vis));
queue<Pos> Q;
Q.push((Pos){sx, sy, 1});
vis[sx][sy] = true;
while(!Q.empty()) {
Pos p = Q.front(); Q.pop();
if(p.x == ex && p.y == ey) return p.s;
Pos np;
for(int d=0; d<4; d++) {
np.x = p.x + dx[d];
np.y = p.y + dy[d];
np.s = p.s + 1;
if(np.x < 0 || np.x > n || np.y < 0 || np.y > m) continue;
if(vis[np.x][np.y]) continue;
if(G[np.x][np.y] != '#') {
vis[np.x][np.y] = true;
Q.push(np);
}
}
}
return -1;
}
int main() {
int T, d1, d2;
//freopen("my.txt", "r", stdin);
scanf("%d", &T);
while(T--) {
scanf("%d%d", &m, &n);
for(int i=0; i<n; i++) {
scanf("%s", G[i]);
for(int j=0; j<m; j++) {
if(G[i][j] == 'S') { sx = i; sy = j; }
else if(G[i][j] == 'E') { ex = i; ey = j; }
}
}
if(sx == 0) { d1 = 3; d2 = 3; }
else if(sx == n-1) { d1 = 1; d2 = 1; }
else if(sy == 0) { d1 = 2; d2 = 0; }
else if(sy == m-1) { d1 = 0; d2 = 2; }
printf("%d ", dfs(sx, sy, d1, 1, dl));
printf("%d ", dfs(sx, sy, d2, 1, dr));
printf("%d\n", BFS(sx, sy));
}
return 0;
}
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
string c[45];
int ans=0,n,num=10000,m;
struct node
{
int x;
int y;
} s,e;
int vis[45][45],Map[45][45];
int count[50][50];
int fx[]= {0,0,-1,1};
int fy[]= {-1,1,0,0};
int DFS(int x,int y)
{
if(x==e.x&&y==e.y)
{
return 0;
}
for(int i=0; i<4; i++)
{
int xx=x+fx[i];
int yy=y+fy[i];
if(xx<m&&yy<n&&xx>=0&&yy>=0&&vis[xx][yy]==0&&Map[xx][yy]&&count[x][y]+1<count[xx][yy])
{
vis[xx][yy]=1;
count[xx][yy]=count[x][y]+1;
DFS(xx,yy);
vis[xx][yy]=0;
}
}
return 0;
}
int Lvis[1606],Rvis[1606];
int Lnum,Rnum;//1上2下3左4右
int Left(int x,int y)
{
while((x!=e.x||y!=e.y))
{
if(Lvis[Lnum]==1)
{
if(Map[x][y-1])
{
y--;
Lvis[++Lnum]=3;
}
else if(Map[x-1][y])
{
x--;
Lvis[++Lnum]=1;
}
else if(Map[x][y+1])
{
y++;
Lvis[++Lnum]=4;
}
else if(Map[x+1][y])
{
x++;
Lvis[++Lnum]=2;
}
continue;
}
if(Lvis[Lnum]==2)
{
if(Map[x][y+1])
{
y++;
Lvis[++Lnum]=4;
}
else if(Map[x+1][y])
{
x++;
Lvis[++Lnum]=2;
}
else if(Map[x][y-1])
{
y--;
Lvis[++Lnum]=3;
}
else if(Map[x-1][y])
{
x--;
Lvis[++Lnum]=1;
}
continue;
}
if(Lvis[Lnum]==3)
{
if(Map[x+1][y])
{
x++;
Lvis[++Lnum]=2;
}
else if(Map[x][y-1])
{
y--;
Lvis[++Lnum]=3;
}
else if(Map[x-1][y])
{
x--;
Lvis[++Lnum]=1;
}
else if(Map[x][y+1])
{
y++;
Lvis[++Lnum]=4;
}
continue;
}
if(Lvis[Lnum]==4)
{
if(Map[x-1][y])
{
x--;
Lvis[++Lnum]=1;
}
else if(Map[x][y+1])
{
y++;
Lvis[++Lnum]=4;
}
else if(Map[x+1][y])
{
x++;
Lvis[++Lnum]=2;
}
else if(Map[x][y-1])
{
y--;
Lvis[++Lnum]=3;
}
continue;
}
}
}
int Right(int x,int y)
{
while(x!=e.x||y!=e.y)
{
if(Rvis[Rnum]==1)
{
if(Map[x][y+1])
{
y++;
Rvis[++Rnum]=4;
}
else if(Map[x-1][y])
{
x--;
Rvis[++Rnum]=1;
}
else if(Map[x][y-1])
{
y--;
Rvis[++Rnum]=3;
}
else if(Map[x+1][y])
{
x++;
Rvis[++Rnum]=2;
}
continue;
}
if(Rvis[Rnum]==2)
{
if(Map[x][y-1])
{
y--;
Rvis[++Rnum]=3;
}
else if(Map[x+1][y])
{
x++;
Rvis[++Rnum]=2;
}
else if(Map[x][y+1])
{
y++;
Rvis[++Rnum]=4;
}
else if(Map[x-1][y])
{
x--;
Rvis[++Rnum]=1;
}
continue;
}
if(Rvis[Rnum]==3)
{
if(Map[x-1][y])
{
x--;
Rvis[++Rnum]=1;
}
else if(Map[x][y-1])
{
y--;
Rvis[++Rnum]=3;
}
else if(Map[x+1][y])
{
x++;
Rvis[++Rnum]=2;
}
else if(Map[x][y+1])
{
y++;
Rvis[++Rnum]=4;
}
continue;
}
if(Rvis[Rnum]==4)
{
if(Map[x+1][y])
{
x++;
Rvis[++Rnum]=2;
}
else if(Map[x][y+1])
{
y++;
Rvis[++Rnum]=4;
}
else if(Map[x-1][y])
{
x--;
Rvis[++Rnum]=1;
}
else if(Map[x][y-1])
{
y--;
Rvis[++Rnum]=3;
}
continue;
}
}
}
int main()
{
int T;
while(cin>>T)
{
while(T--)
{
num=0x3f3f3f3f;
Lnum=1;
Rnum=1;
cin>>n>>m;
for(int i=0; i<m; i++)
cin>>c[i];
for(int i=0; i<m; i++)
for(int j=0; j<n; j++)
{
if(c[i][j]=='#')
Map[i][j]=0;
else
Map[i][j]=1;
if(c[i][j]=='S')
{
s.x=i;
s.y=j;
}
if(c[i][j]=='E')
{
e.x=i;
e.y=j;
}
}
if(s.x==0)
{
Lvis[Lnum]=2;
Rvis[Rnum]=2;
}
else if(s.x==m-1)
{
Lvis[Lnum]=1;
Rvis[Rnum]=1;
}
else if(s.y==0)
{
Lvis[Lnum]=4;
Rvis[Rnum]=4;
}
else if(s.y==n-1)
{
Lvis[Lnum]=3;
Rvis[Rnum]=3;
}
for(int i=0;i<m;i++)
for(int j=0;j<n;j++)
count[i][j]=10000000;
count[s.x][s.y]=1;
vis[s.x][s.y]=1;
memset(vis,0,sizeof(vis));
DFS(s.x,s.y);
Left(s.x,s.y);
Right(s.x,s.y);
cout<<Lnum<<" "<<Rnum<<" "<<count[e.x][e.y]<<endl;
}
}
}