HDU 1242 Rescue 【DFS】
2014-12-08 13:37
351 查看
Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17082 Accepted Submission(s): 6185
Problem Description
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
/*
题解:简单搜索题,DFS可过。
*/
#include<cstdio> #include<cstring> const int N = 202; const int inf = 0x3f3f3f3f; int n,m,sx,sy,tot,ans,ok; int vis ,dir[4][2]={1,0,-1,0,0,1,0,-1}; char map ; void dfs(int x,int y,int step) { if(map[x][y]=='r') { ok=1; if(step<ans) ans=step; return; } vis[x][y]=1; for(int i=0; i<4; i++) { int a=x+dir[i][0]; int b=y+dir[i][1]; if(a>=0&&a<n&&b>=0&&b<m&&!vis[a][b]&&map[a][b]!='#') { if(map[a][b]=='x') { dfs(a,b,step+2); } if(map[a][b]=='.'||map[a][b]=='r')//原来这一步没看清楚,倒弄了好长时间 { dfs(a,b,step+1); } } } vis[x][y]=0; } int main(){ while(scanf("%d %d",&n,&m)!=EOF) { tot=ok=0,ans=inf; for(int i=0; i<n; i++) { scanf("%s",map[i]); for(int j=0; j<m; j++) { if(map[i][j]=='a') { sx=i,sy=j; } } } map[sx][sy]='#'; memset(vis,0,sizeof(vis)); dfs(sx,sy,0); if(ok) printf("%d\n",ans); else printf("Poor ANGEL has to stay in the prison all his life.\n"); } return 0; }
相关文章推荐
- HDU 1242 Rescue (搜索 DFS)
- HDU 1242 Rescue (BFS or DFS)
- HDU 1242 Rescue(DFS or BFS)
- hdu 1242Rescue(DFS)
- hdu1242 Rescue (dfs+剪枝)
- 模板:HDU 1242: Rescue: (BFS DFS)
- HDU 1242 Rescue(DFS)
- hdu 1242 Rescue(方法一:BFS+优先队列,方法二:DFS)
- HDU 1242 Rescue(DFS入门)
- HDU 1242 -- Rescue (bfs,dfs)
- 对BFS与DFS的一些看法 ( 以HDU 1242 Rescue为例 )
- HDU 1242 Rescue - DFS 回溯
- HDU 1242 Rescue(dfs深搜)
- HDU 1242 Rescue (DFS)
- HDU 1242 Rescue (DFS+剪枝,一个起点多个终点)
- [ACM] hdu 1242 Rescue (BFS+优先队列)
- BFS:HDU-1242-Rescue(带守卫的迷宫问题)(优先队列)
- hdu 1242 zoj1649 Rescue
- hdu 1242 ( Rescue )
- HDU 1242 Rescue(营救)(优先队列+BFS)