Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27406 Accepted Submission(s): 9711
was caught by the MOLIGPY! He was put in * by Moligpy. The *
is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs,
and GUARDs in the *.
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.)
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.
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 * all his
life."
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........
//一道很简单的题卡了很长时间。从重点出发找起点，优先队列存
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<functional>
using namespace std;
int n,m;
struct Lu
{
int x,y,cnt;
friend bool operator<(Lu a,Lu b)
{
return a.cnt>b.cnt;
}
};
bool vis[][];
int dir[][]={,,-,,,,,-};
char ch[][];
int bfs(int sx,int sy)
{
priority_queue<Lu>q;
Lu L1,L2;
L1.x=sx;L1.y=sy;L1.cnt=;
q.push(L1);
vis[sx][sy]=;
while(!q.empty())
{
L2=q.top();
q.pop();
if(ch[L2.x][L2.y]=='r')
return L2.cnt;
for(int i=;i<;i++)
{
L1.x=L2.x+dir[i][];
L1.y=L2.y+dir[i][];
if(L1.x<||L1.x>=n||L1.y<||L1.y>=m) continue;
if(vis[L1.x][L1.y]) continue;
if(ch[L1.x][L1.y]=='#') continue;
if(ch[L1.x][L1.y]=='x')
L1.cnt=L2.cnt+;
else L1.cnt=L2.cnt+;
vis[L1.x][L1.y]=;
q.push(L1);
}
}
return -;
}
int main()
{
int sx,sy;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=;i<n;i++)
{
scanf("%s",ch[i]);
for(int j=;j<m;j++)
{
if(ch[i][j]=='a')
{
sx=i;sy=j;
}
}
}
memset(vis,,sizeof(vis));
int ans=bfs(sx,sy);
if(ans==-) printf("Poor ANGEL has to stay in the * all his life.\n");
else printf("%d\n",ans);
}
return ;
}