hdu 2822 bfs

问题可以抽象为求起点到终点的最短路，其中经过x的代价是0，而经过.的代价是1，bfs的时候用优先队列即可。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;

const int N = 1001;
char maze[N][N];
bool visit[N][N];
int dx[] = { 0, 0, 1, -1 };
int dy[] = { 1, -1, 0, 0 };
int n, m, sx, sy, ex, ey;

bool ok( int x, int y )
{
    return x >= 0 && x < n && y >= 0 && y < m && !visit[x][y];
}

struct Node 
{
    int x, y, t;
    Node(){}
    Node( int _x, int _y, int _t )
    {
        x = _x, y = _y, t = _t;
    }
    bool operator < ( const Node & o ) const 
    {
        return t > o.t;
    }
};

priority_queue<Node> q;

int bfs()
{
    while ( !q.empty() ) q.pop();
    memset( visit, 0, sizeof(visit) );
    q.push( Node( sx, sy, 0 ) );
    visit[sx][sy] = 1;
    while ( !q.empty() ) 
    {
        Node cur = q.top();
        q.pop();
        if ( cur.x == ex && cur.y == ey ) return cur.t;
        for ( int i = 0; i < 4; i++ )
        {
            int x = cur.x + dx[i];
            int y = cur.y + dy[i];
            if ( ok( x, y ) )
            {
                int d = maze[x][y] == '.' ? 1 : 0;
                q.push( Node( x, y, cur.t + d ) );
                visit[x][y] = 1;
            }
        }
    }
    return -1;
}

int main ()
{
    while ( scanf("%d%d", &n, &m) != EOF )
    {
        if ( !n && !m ) break;
        for ( int i = 0; i < n; i++ )
        {
            scanf("%s", maze[i]);
        }
        scanf("%d%d%d%d", &sx, &sy, &ex, &ey);
        sx--, sy--, ex--, ey--;
        int ans = bfs();
        printf("%d
", ans);
    }
    return 0;
}