Rescue--hdu1242

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21431    Accepted Submission(s): 7641

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

 

这个题是用广搜写的！但是这个题是有点特殊的，因为遇到X的时候时间是要再加一的！所以可以用优先队列来解决，但是还是可以用普通队列来解决的，方法就是遇到X先加一，然后标记走过，再后来再遇到X直接加一，不再往四周搜索！等于走了两步！详情如下

普通队列版：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define ma 210
using namespace std;
char map[ma][ma];
int v[ma][ma],m,n;
int mov[4][2]={1,0,-1,0,0,1,0,-1};
struct node
{
    int x,y,step,flag;
};
bool can(node x)
{
    if(x.x>=0&&x.x<m&&x.y>=0&&x.y<n&&(map[x.x][x.y]!='#')&&!v[x.x][x.y])
    return true ;
    return false;
}
int bfs(int x,int y)
{
    int i;
    memset(v,0,sizeof(v));
    queue<node>q;
    node ta,tb;
    ta.x=x;
    ta.y=y;
    ta.step=0;
    ta.flag=0;
    q.push(ta);
    while(!q.empty())
    {
        ta=q.front();
        q.pop();
        if(ta.flag==1)
        {
            ta.step++;
            ta.flag=0;
            q.push(ta);
            continue;
        }//第二次遇到时，flag已经等于1，这时候不能走了，步数要加一，而且flag要重新标记为0
        if(map[ta.x][ta.y]=='a')
            return ta.step;
        for(i=0;i<4;i++)
        {
            tb.x=ta.x+mov[i][0];
            tb.y=ta.y+mov[i][1];
            tb.step=ta.step+1;
            if(can(tb))
            {

                if(map[tb.x][tb.y]=='x')
                tb.flag=1;
                else
                tb.flag=0;//第一次遇到X就标记为flag=1，否则为0！
                v[tb.x][tb.y]=1;
                q.push(tb);
            }
        }
    }
    return -1;    
}
int main()
{
    int i,j,a,b;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        getchar();
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='r')
                {
                    a=i;
                    b=j;
                }
            }
            getchar();
        }
        v[a][b]=1;
        int rr=bfs(a,b);
        if(rr==-1)
        printf("Poor ANGEL has to stay in the prison all his life.
");
        else
        printf("%d
",rr);
    }
    return 0;
 } 

优先队列版：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define maxn 210
using namespace std;
int m,n,v[maxn][maxn],mov[4][2]={1,0,-1,0,0,1,0,-1};
char map[maxn][maxn];

struct node
{
    int x,y,step;
    friend bool operator <(node x,node y)
    {
        return x.step>y.step;
    }
};
priority_queue<node>q;
bool can(node x)
{
    if(!v[x.x][x.y]&&x.x>=0&&x.x<m&&x.y>=0&&x.y<n&&map[x.x][x.y]!='#')//(map[x.x][x.y]=='.'||map[x.x][x.y]=='x'))
    return  true;
    return false;
 } 
int bfs(int x,int y)
{
    int i;
    node ta,tb;
    ta.x=x;
    ta.y=y;
    ta.step=0;
    q.push(ta);
    while(!q.empty())
    {
        ta=q.top();
        q.pop();
        if(map[ta.x][ta.y]=='a')
            return ta.step;//到终点就返回队首，用的优先队列所以队首的步数最少！
        for(i=0;i<4;i++)
        {
            tb.x=ta.x+mov[i][0];
            tb.y=ta.y+mov[i][1];
            if(can(tb))
            {
                if(map[tb.x][tb.y]=='x')
                tb.step=ta.step+2;//遇到X一定要加2
                else
                tb.step=ta.step+1;//其他加1
                v[tb.x][tb.y]=1;
                q.push(tb);
            }

        }
    }
    return -1;//找不到就返回-1
}
int main()
{
    int i,j,a,b;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        getchar();
        memset(v,0,sizeof(v));
        for(i=0;i<m;i++)
        {
            for(j=0;j<n;j++)
            {
                scanf("%c",&map[i][j]);
                if(map[i][j]=='r')
                a=i,b=j;
            }
            getchar();
        }
        v[a][b]=1;
        int rr=bfs(a,b);
        if(rr==-1)
        printf("Poor ANGEL has to stay in the prison all his life.
");
        else
        printf("%d
",rr);
    }
    return 0;
}