u013044116 于 2014.10.01 20:19 提问

ACM一道题 poj3523 UVA1601双向广度优先BFS

1个回答

lxj1048580178   2015.01.17 14:57

#include
#include
#include
#include
using namespace std;

const int maxs = 20;
const int maxn = 150; // 75% cells plus 2 fake nodes
const int dx[]= {1,-1,0,0,0}; // 4 moves, plus "no move"
const int dy[]= {0,0,1,-1,0};

inline int ID(int a, int b, int c)
{
return (a<<16)|(b<<8)|c;
}

int s[3], t[3]; // starting/ending position of each ghost

int deg[maxn], G[maxn][5]; // target cells for each move (including "no move")

inline bool conflict(int a, int b, int a2, int b2)
{
return a2 == b2 || (a2 == b && b2 == a);
}

int d[maxn][maxn][maxn]; // distance from starting state

int bfs()
{
queue q;
memset(d, -1, sizeof(d));
q.push(ID(s[0], s[1], s[2])); // starting node
d[s[0]][s[1]][s[2]] = 0;
while(!q.empty())
{
int u = q.front();
q.pop();
int a = (u>>16)&0xff, b = (u>>8)&0xff, c = u&0xff;
if(a == t[0] && b == t[1] && c == t[2]) return d[a][b][c]; // solution found
for(int i = 0; i < deg[a]; i++)
{
int a2 = G[a][i];
for(int j = 0; j < deg[b]; j++)
{
int b2 = G[b][j];
if(conflict(a, b, a2, b2)) continue;
for(int k = 0; k < deg[c]; k++)
{
int c2 = G[c][k];
if(conflict(a, c, a2, c2)) continue;
if(conflict(b, c, b2, c2)) continue;
if(d[a2][b2][c2] != -1) continue;
d[a2][b2][c2] = d[a][b][c]+1;
q.push(ID(a2, b2, c2));
}
}
}
}
return -1;
}

int main()
{
int w, h, n;

``````while(scanf("%d%d%d\n", &w, &h, &n) == 3 && n)
{
char maze[20][20];
for(int i = 0; i < h; i++)
fgets(maze[i], 20, stdin);

// extract empty cells
int cnt, x[maxn], y[maxn], id[maxs][maxs]; // cnt is the number of empty cells
cnt = 0;
for(int i = 0; i < h; i++)
for(int j = 0; j < w; j++)
if(maze[i][j] != '#')
{
x[cnt] = i;
y[cnt] = j;
id[i][j] = cnt;
if(islower(maze[i][j])) s[maze[i][j] - 'a'] = cnt;
else if(isupper(maze[i][j])) t[maze[i][j] - 'A'] = cnt;
cnt++;
}

// build a graph of empty cells
for(int i = 0; i < cnt; i++)
{
deg[i] = 0;
for(int dir = 0; dir < 5; dir++)
{
int nx = x[i]+dx[dir], ny = y[i]+dy[dir];
// "Outermost cells of a map are walls" means we don't need to check out-of-bound
if(maze[nx][ny] != '#') G[i][deg[i]++] = id[nx][ny];
}
}

// add fakes nodes so that in each case we have 3 ghosts. this makes the code shorter
if(n <= 2)
{
deg[cnt] = 1;
G[cnt][0] = cnt;
s[2] = t[2] = cnt++;
}
if(n <= 1)
{
deg[cnt] = 1;
G[cnt][0] = cnt;
s[1] = t[1] = cnt++;
}

printf("%d\n", bfs());
}
return 0;
``````

}