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
廣度優先遍歷求解。 不同於一般的迷宮最短路徑求解，每步的時間是不確定的。因此需要用一個附加數據保存時間，並遍歷所有可達的情況。
那麼BFS是否會無限搜索下去呢？不會的，因為我們加入了判斷條件是，這種走法比之前所花費的時間更少。
[cpp]
#include <iostream>
#include <queue>
#include <stdio.h>
using namespace std;
#define MAXMN 200
#define INF 1000000
class Point{
public:
int x,y,time;
};
queue<Point> Q;
int N,M;
char map[MAXMN][MAXMN];
int mintime[MAXMN][MAXMN]; //表示到達當前位置所需要的 最短時間
int dir[4][2] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };
int ax,ay; //結束位置
int bfs(Point s){ //從s點開始搜索
Q.push(s);
Point hd;
while( !Q.empty()){
hd = Q.front();
Q.pop();
for(int i=0; i<4; i++){
int x = hd.x + dir[i][0];
int y = hd.y + dir[i][1];
if(x >= 0 && x <N && y >=0 && y < M && map[x][y] != '#'){
Point t;
t.x = x;
t.y = y;
t.time = hd.time + 1;
if(map[x][y] == 'x') t.time ++;
if(t.time < mintime[x][y]){
mintime[x][y] = t.time;
Q.push(t);
}
}
}
}
return mintime[ax][ay];
}
int main() {
while(scanf("%d %d", &N,&M) != EOF){
for(int i=0; i<N; i++)
scanf("%s", map[i]);
int sx,sy;
Point start;
for(int i=0; i<N; i++){
for(int j=0; j<M; j++){
mintime[i][j] = INF; //mintime初始為最大，不可達
if(map[i][j] == 'a'){
ax = i;
ay = j;
}else if(map[i][j] == 'r'){
sx = i;
sy = j;
}
}
}
start.x = sx;
start.y = sy;
start.time = 0;
mintime[sx][sy] = 0;
int mint = bfs(start);
if(mint < INF)
printf("%d\n",mint);
else
printf("Poor ANGEL has to stay in the prison all his life.\n");
}
return 0;
}
#include <iostream>
#include <queue>
#include <stdio.h>
using namespace std;
#define MAXMN 200
#define INF 1000000
class Point{
public:
int x,y,time;
};
queue<Point> Q;
int N,M;
char map[MAXMN][MAXMN];
int mintime[MAXMN][MAXMN]; //表示到達當前位置所需要的 最短時間
int dir[4][2] = { { -1, 0 }, { 0, -1 }, { 1, 0 }, { 0, 1 } };
int ax,ay; //結束位置
int bfs(Point s){ //從s點開始搜索
Q.push(s);
Point hd;
while( !Q.empty()){
hd = Q.front();
Q.pop();
for(int i=0; i<4; i++){
int x = hd.x + dir[i][0];
int y = hd.y + dir[i][1];
if(x >= 0 && x <N && y >=0 && y < M && map[x][y] != '#'){
Point t;
t.x = x;
t.y = y;
t.time = hd.time + 1;
if(map[x][y] == 'x') t.time ++;
if(t.time < mintime[x][y]){
mintime[x][y] = t.time;
Q.push(t);
}
}
}
}
return mintime[ax][ay];
}
int main() {
while(scanf("%d %d", &N,&M) != EOF){
for(int i=0; i<N; i++)
scanf("%s", map[i]);
int sx,sy;
Point start;
for(int i=0; i<N; i++){
for(int j=0; j<M; j++){
mintime[i][j] = INF; //mintime初始為最大，不可達
if(map[i][j] == 'a'){
ax = i;
ay = j;
}else if(map[i][j] == 'r'){
sx = i;
sy = j;
}
}
}
start.x = sx;
start.y = sy;
start.time = 0;
mintime[sx][sy] = 0;
int mint = bfs(start);
if(mint < INF)
printf("%d\n",mint);
else
printf("Poor ANGEL has to stay in the prison all his life.\n");
}
return 0;
}
