hdu 4198 Quick out of the Harbour（BFS+优先队列）

题目链接：hdu4198

题目大意：求起点S到出口的最短花费，其中#为障碍物，无法通过，‘.’的花费为1 ，@的花费为d+1。

需注意起点S可能就是出口，因为没考虑到这个，导致WA很多次.......

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;
char map[505][505];
int d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int n,m,t;
int begin_x,begin_y,end_x,end_y;
struct node
{
    int x,y,time;
    friend bool operator < (node a,node b)
    {
        return a.time > b.time;
    }
};
void bfs()
{
    priority_queue <node> q;
    node s,temp;
    s.x = begin_x;
    s.y = begin_y;
    s.time = 0;
    map[begin_x][begin_y] = '#';
    q.push(s);
    while(!q.empty())
    {
        temp = q.top();
        q.pop();
        if(temp.x == end_x && temp.y == end_y)
        {
            printf("%d
",temp.time + 1);
            return;
        }
        for(int i = 0 ; i < 4 ; i ++)
        {
            s.x = temp.x + d[i][0];
            s.y = temp.y + d[i][1];
            if(s.x < 0 || s.x >= n || s.y < 0 || s.y >= m || map[s.x][s.y] == '#')
            continue;
            if(map[s.x][s.y] == '.') s.time = temp.time + 1;
            else s.time = temp.time + t + 1;
            map[s.x][s.y] = '#';
            q.push(s);
        }
    }
}
int main()
{
    int T,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&t);
        for(i = 0 ; i < n ; i ++)
        {
            scanf("%s",map[i]);
            for(j = 0 ; j < m ; j ++)
            {
                if(map[i][j] == 'S')
                {
                    begin_x = i;
                    begin_y = j;
                }
                if( (i == 0 || i == n - 1 || j == 0 || j == m - 1) && map[i][j] != '#')//刚开始用的else if，没有考虑起点也是终点的情况，WA了很多次
                {
                    end_x = i;
                    end_y = j;
                }
            }
        }
        bfs();
    }
    return 0;
}