poj1376 搜索(BFS)

题目: http://poj.org/problem?id=1376
题意: 可以给机器人发两种指令：(1) GO： 一次朝一个方向移动1或2或3步 (2) TRUE： 转左或转右(一次只能转90度，不能连转两次) 。 每次执行1个指令都要花1秒。而且边界不能走。

思路： 用BFS，有三维数组保存状态，代表到达当前位置（x,y)在z方向上花的秒数(权重), 比较权重。比之前的状态权重小的，则覆盖当前状态。

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int  map[60][60];
int MinTime[60][60][4];//保存状态
int B1,B2,E1,E2;
char TRUE[10];
int direction[4][2] = {{0,1},{1,0},{0,-1},{-1,0}};
int M,N;
struct cnode
{
int x;
int y;
int dir;//当前的方向
int cost;//当前花的时间到达这里
};
int  bfs()
{
int i;
queue<cnode> qu;
cnode node;
node.x = B1;
node.y = B2;
if(map[B1][B2] != 0 || map[E1][E2] !=0 )
{
return -1;
}
node.cost = 0;
if(strcmp(TRUE,"east") == 0)
{
node.dir = 0;
}
else if(strcmp(TRUE,"south") == 0)
{
node.dir = 1;
}
else if(strcmp(TRUE, "west") == 0)
{
node.dir = 2;
}
else
{
node.dir = 3;
}
memset(MinTime, 127, sizeof(MinTime));
qu.push(node);
cnode current;
MinTime[node.x][node.y][node.dir] = 0;
while(!qu.empty())
{
current = qu.front();
if(current.x == E1 &¤t.y == E2)
{
return current.cost;
}
cnode jump;
qu.pop();
//if(current.dir == 0)
//{
for(i = 1; i <= 3; i++)
{
jump = current;
//			jump.y += i;
jump.x += i * direction[current.dir][0];
jump.y += i * direction[current.dir][1];
jump.cost++;
if(map[jump.x][jump.y] != 0)
{
break;
}

if(map[jump.x][jump.y] == 0 && MinTime[jump.x][jump.y][jump.dir] > jump.cost)
{
MinTime[jump.x][jump.y][jump.dir] = jump.cost;
qu.push(jump);
}
}
/*			for(i = 1; i < 4; i++)
{
jump = current;
jump.dir += i;
jump.dir %= 4;
int flag =  abs(jump.dir - current.dir);
if(flag == 3)
{
jump.cost += 1;
}
else
{
jump.cost += flag;
}
//if(map[jump.x][jump.y] == 0 && HasTrue[jump.x][jump.y] == false)
if(map[jump.x][jump.y] == 0 && MinTime[jump.x][jump.y][jump.dir] > jump.cost)
{
MinTime[jump.x][jump.y][jump.dir] = jump.cost;

qu.push(jump);
}

}*/

//		HasTrue[current.x][current.y] = true;
//////下面是转方向的
jump = current;
jump.dir += 1;
jump.dir %= 4;
jump.cost++;
if(map[jump.x][jump.y] == 0 &&  MinTime[jump.x][jump.y][jump.dir] > jump.cost)
{
MinTime[jump.x][jump.y][jump.dir] = jump.cost;

qu.push(jump);
}
jump = current;
jump.dir -= 1;
jump.dir %= 4;
if(jump.dir < 0)
{
jump.dir += 4;
}
jump.cost++;
if(map[jump.x][jump.y] == 0 && MinTime[jump.x][jump.y][jump.dir] > jump.cost)
{
MinTime[jump.x][jump.y][jump.dir] = jump.cost;

qu.push(jump);
}
}
return -1;
}
int main()
{
//	freopen("poj1376.in", "r", stdin);
//freopen("g.in", "r", stdin);
//freopen("poj1376.out","w",stdout);
while(scanf("%d %d",&M,&N) != EOF)
{
if(M == 0 && N == 0)
{
break;
}
int  input[M + 10][N + 10];
int i,j;
for(i = 1; i <= M; i++)
{
for(j = 1; j <= N; j++)
{
scanf("%d",&input[i][j]);
}
}
memset(map,255,sizeof(map));
for(i = 1; i <= M + 1; i++)
{
for(j = 1; j <= N + 1; j++)
{
map[i][j] = 0;
}
}
for(i = 1; i <= M; i++)
{
for(j = 1; j <= N; j++)
{
if(input[i][j] == 1)
{
map[i][j] = 1;
map[i][j + 1] = 1;
map[i + 1][j] = 1;
map[i + 1] [j + 1] = 1;
}
}
}
for(i = 1; i <= M + 1; i++)
{
map[i][1] = - 1;
map[i][N + 1] = -1;
}
for(i = 1; i <= N + 1; i++)
{
map[M + 1][i] = -1;
map[1][i] = -1;
}
scanf("%d %d  %d  %d",&B1,&B2,&E1,&E2);
B1++;
B2++;
E1++;
E2++;
scanf("%s",TRUE);
int ans;
ans = bfs();
printf("%d\n",ans);
}
return 0;
}