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\n",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;
}
