【記憶化DFS】HDOJ1242 Rescue

【題目】

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21820    Accepted Submission(s): 7774

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

 

Author

CHEN, Xue

 

Source

ZOJ Monthly, October 2003

 

Recommend

Eddy   |   We have carefully selected several similar problems for you:  1240 1016 1241 1372 1175 

 

Statistic | Submit | Discuss |Note

【思路】

一開始看數據範圍不大，想直接爆搜，寫了一個裸DFS，TLE，見優化無望，又不想改成BFS，上網搜了相關的記憶化來抄了一下。

對記憶化的認識更深入。

特點是

1.主程對每一個位置都展開搜索。

2.dp數組保存當前位置最好的答案，並不斷更新。

3.如果dp數組答案比當前好，那麼剪枝，否則更新。

【代碼】

#include <iostream>
#include <iomanip>
#include <cstring>

using namespace std;

#define INF 1e9

int n,m,dp[205][205],ex,ey,ans;
char ma[205][205];
bool vis[205][205];
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};

void dfs(int x,int y)
{
	if(dp[x][y]>=dp[ex][ey]) return;
	for(int i=0;i<4;++i)
	{
		int xx=x+dx[i];
		int yy=y+dy[i];
		if(ma[xx][yy]=='#'||vis[xx][yy]) continue;//這裏錯寫成break了調了半天T_T
		if(ma[xx][yy]=='x')
		{
			if(dp[xx][yy]==-1||dp[xx][yy]>dp[x][y]+2)
				dp[xx][yy]=dp[x][y]+2;
			else continue;
		}
		else if(ma[xx][yy]=='a')
		{
			if(dp[xx][yy]==-1||dp[xx][yy]>dp[x][y]+1)
				dp[xx][yy]=dp[x][y]+1;
			return;
		}
		else if(ma[xx][yy]=='.')
		{
			if(dp[xx][yy]==-1||dp[xx][yy]>dp[x][y]+1)
				dp[xx][yy]=dp[x][y]+1;
			else continue;
		}
		vis[xx][yy]=true;
		dfs(xx,yy);
		vis[xx][yy]=false;
	}
}

int main()
{
	ios::sync_with_stdio(false);
	int sx,sy;
	while(cin>>n>>m)
	{
		memset(ma,'#',sizeof(ma));
		memset(vis,0,sizeof(vis));
		ex=ey=-1;
		for(int i=1;i<=n;++i)
			for(int j=1;j<=m;++j)
			{
				dp[i][j]=-1;
				cin>>ma[i][j];
				if(ma[i][j]=='r')
				{
					sx=i;
					sy=j;
				}
				if(ma[i][j]=='a')
				{
					ex=i;
					ey=j;
					dp[i][j]=INF;
				}
			}
		if(ex==-01&&ey==-1)
		{
			cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
			continue;
		}
		dp[sx][sy]=0;
		dfs(sx,sy);
		if(dp[ex][ey]!=INF)
			cout<<dp[ex][ey]<<endl;
		else
			cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
	}
	//system("pause");
	return 0;
}