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;
}