杭电 1242 Rescue（广搜）

http://acm.hdu.edu.cn/showproblem.php?pid=1242

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15597    Accepted Submission(s): 5663

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

 

刚开始看到这个题的时候觉得用模板就行啦，不算难，可是敲出代码提交一直错，真心一下子想不通，想我这样脾气暴躁的人，简直要被气死了。

AC代码（一）：用广搜模板做的，侥幸AC。

#include
#include
#include
#include
using namespace std;
#define max 201

char a[max][max];
int move[4][2]={-1,0,1,0,0,1,0,-1},v[max][max];//移动方向排列在这里很重要。
int n,m,ans;
struct node
{
    int x,y,step;
};

void bfs(int i,int j)
{
    queueq;
    node now,next;
    now.x=i;
    now.y=j;
    v[now.x][now.y]=1;
    now.step=0;
    q.push(now);    
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(a[now.x][now.y]=='r')
        {
            ans=now.step;
            return ;
        }
        else
        {
            for(int t=0;t<4;t++)
            {
                next.x=now.x+move[t][0];
                next.y=now.y+move[t][1];
                if(!v[next.x][next.y]&&a[next.x][next.y]!='#'&&next.x>=0&&next.x=0&&next.y>n>>m)
    {
        ans=0;
        memset(v,0,sizeof(v));
        for(i=0;i>a[i][j];
            }
        }
        for(i=0;i


AC代码（二）：用优先队列做的，比上一个靠谱些。

#include
#include
#include
using namespace std;

#define M 205
char c[M][M];
int v[M][M];
int w[4][2]={-1,0,0,-1,0,1,1,0};
int ans,n,m;

struct node 
{
	int x,y,time;
	friend bool operator< (const node a,const node b)
	{
		return a.time>b.time;
	}
};

void bfs(int a,int b)
{
	node now,tmp;
	priority_queue   q;
	now.time=0;
	now.x=a;
	now.y=b;
	memset(v,0,sizeof(v));
	v[a][b]=1;
	q.push(now);

	while(!q.empty())
	{
		now=q.top();
		q.pop();
		if(c[now.x][now.y]=='r')
		{
			ans=now.time;
			return ;
		}

		for(int i=0;i<4;i++)
		{
			tmp.x=now.x+w[i][0];
			tmp.y=now.y+w[i][1];
			if(tmp.x>=0&&tmp.x=0&&tmp.y>n>>m)
	{
		int i,j,x,y;
		for(i=0;i>c[i][j];
				if(c[i][j]=='a')
					x=i,y=j;
			}

		ans=0;
		bfs(x,y);
		if(ans)
			cout<