Rescue
Time Limit: 2 Seconds Memory Limit: 65536 KB
Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.
Angel‘s friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there‘s a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
Input
First line contains two integers stand for N and M.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel‘s friend.
Process to the end of the file.
Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."
Sample Input
7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........
Sample Output
13
如题。简单的广搜水题，问题在于如何扩展x的结点。方法也很简单。可以把x看做一个有两个结点的连通分量，第一次进入x结点时，只能扩展出一个结点（还是x自己），第二次进入x结点时，才能扩展出周围四个方向的结点。
当然，为了避免出现死循环，必须要把x结点的值更改。我是更改为‘.‘。
最后的最小值，则是广搜树的高度。
下面是AC代码
#include <stdio.h>
#include <queue>
using namespace std;
/*
整整两个小时的时间！！！！
居然是忘记清空队列的低级BUG！！！！
1、标记是否访问过某个结点，如果访问过，则不再入队；
2、如果当前的结点值是x，则需要将这个访问过的结点再次入队，并更改为.，本轮不再扩展。类似于将该点展开为一个线段，头扩展的结点是尾，尾扩展周围的结点。
*/
typedef struct point
{
int x;
int y;
}point;
queue<point> road;
char prison[201][201];
int visit[201][201];
int high[201][201];
int n,m;
int startX,startY;
void init()
{
for(int i = 0;i < n;i++)
{
for(int j = 0;j < m;j++)
{
if(prison[i][j] == ‘a‘)
{
startX = i;
startY = j;
}
}
}
for(int i = 0;i < 201;i++)
for(int j = 0;j < 201;j++)
{
high[i][j] = 0;
visit[i][j] = 0;
}
while(!road.empty())
road.pop(); //f*cking bug occurs here.
}
void BFS()
{
point tmp;
point p1;
tmp.x = startX;
tmp.y = startY;
int finalX,finalY;
high[startX][startY] = 0;
visit[startX][startY] = 1;
road.push(tmp);
int flag = 0;
while(!road.empty())
{
tmp = road.front();
road.pop();
if(prison[tmp.x][tmp.y] == ‘x‘)
{
road.push(tmp);
high[tmp.x][tmp.y] ++;
prison[tmp.x][tmp.y] = ‘.‘;
}
else if(prison[tmp.x][tmp.y] == ‘r‘)
{
flag = 1;
finalX = tmp.x;
finalY = tmp.y;
break;
}
else
{
for(int i = -1;i <= 1;i++)
{
if(0 == i)continue;
if(tmp.x + i >= 0 && tmp.x + i < n)
{
if(visit[tmp.x+i][tmp.y] == 0 && prison[tmp.x+i][tmp.y] != ‘#‘)
{
visit[tmp.x+i][tmp.y] = 1;
p1.x = tmp.x + i;
p1.y = tmp.y;
high[p1.x][p1.y] = high[tmp.x][tmp.y] + 1;
road.push(p1);
}
}
if(tmp.y + i >= 0 && tmp.y + i < m)
{
if(visit[tmp.x][tmp.y+i] == 0 && prison[tmp.x][tmp.y+i] != ‘#‘)
{
visit[tmp.x][tmp.y+i] = 1;
p1.x = tmp.x;
p1.y = tmp.y + i;
high[p1.x][p1.y] = high[tmp.x][tmp.y] + 1;
road.push(p1);
}
}
}
}
}
if(flag)
{
printf("%d\n",high[finalX][finalY]);
}
else printf("Poor ANGEL has to stay in the prison all his life.\n");
}
int main(void)
{
while(scanf("%d%d",&n,&m) != EOF)
{
for(int i = 0;i < n;i++)
scanf("%s",prison[i]);
init();
BFS();
}
return 0;
}