hdu 1180(诡异的楼梯)

这题还真是诡异,明明思路很清晰,BFS也很熟练 的敲出来了,可就是改了半天,原来问题在这里,“Harry从来不在楼梯上停留”,这里提示的很明显了,他除了不再楼梯上停留,其他地方是可以停留的,当然,也就只有在遇到楼梯,又暂时过不去时,才有必要在原地停留,郁闷呀,疏忽了这里,wa了好几次……

hdu1180 代码
#include <iostream>
#include
<queue>
using namespace std;
struct Node
{
int x, y, time;
Node(
int _x=0,int _y=0,int _time=0):x(_x),y(_y),time(_time){};
};

Node first;
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int n, m, a, b;
char map[21][21];
bool vis[21][21];
queue
<Node> Q;

int bfs()
{
int fx, fy;
first.x
= a;
first.y
= b;
first.time
= 0;
Q.push(first);
while (!Q.empty())
{
first
= Q.front();
Q.pop();

if (map[first.x][first.y] == 'T')
{
return first.time;
}

int i;
for (i = 0; i < 4; ++i)
{
fx
= first.x + dir[i][0];
fy
= first.y + dir[i][1];

if (fx >= 0 && fx < n && fy >= 0 && fy < m && map[fx][fy] != '*' && !vis[fx][fy])
{
if (map[fx][fy] == '.' || map[fx][fy] == 'T')
{
vis[fx][fy]
= 1;
Q.push(Node(fx,fy,first.time
+1));
}
else if (map[fx][fy] == '-')
{
if (first.time % 2 == 0)
{
if (fx == first.x)
{
fy
= 2*fy - first.y;
if (fy >= 0 && fy < m && map[fx][fy] != '*' && !vis[fx][fy])
{
vis[fx][fy]
= 1;
Q.push(Node(fx,fy,first.time
+1));
}
}
else
Q.push(Node(first.x,first.y,first.time
+1));//遇到楼梯,但方向不对,原地停留
}
else
{
if (fy == first.y)
{
fx
= 2*fx - first.x;
if (fx >= 0 && fx < n && map[fx][fy] != '*' && !vis[fx][fy])
{
vis[fx][fy]
= 1;
Q.push(Node(fx,fy,first.time
+1));
}
}
else
Q.push(Node(first.x,first.y,first.time
+1));//同样,原地停留
}
}
else if (map[fx][fy] == '|')
{
if (first.time % 2 == 0)
{
if (fy == first.y)
{
fx
= 2*fx - first.x;
if (fx >= 0 && fx < n && map[fx][fy] != '*' && !vis[fx][fy])
{
vis[fx][fy]
= 1;
Q.push(Node(fx,fy,first.time
+1));
}
}
else
Q.push(Node(first.x,first.y,first.time
+1));
}
else
{
if (fx == first.x)
{
fy
= 2*fy - first.y;
if (fy >= 0 && fy < m && map[fx][fy] != '*' && !vis[fx][fy])
{
vis[fx][fy]
= 1;
Q.push(Node(fx,fy,first.time
+1));
}
}
else
Q.push(Node(first.x,first.y,first.time
+1));
}
}
}
}
}
return -1;
}


int main()
{
while (scanf("%d %d", &n, &m) != EOF)
{
int i, j;
for (i = 0; i < n; ++i)
{
getchar();
for (j = 0; j < m; ++j)
{
scanf(
"%c", &map[i][j]);
if (map[i][j] == 'S')
{
a
= i;
b
= j;
}
vis[i][j]
=0;
}
}
vis[a][b]
= 1;
while (!Q.empty())
{
Q.pop();
}
printf(
"%d\n",bfs());
}
return 0;
}
原文地址:https://www.cnblogs.com/nanke/p/2126020.html