hdu2612Find a way(dbfs或者IDA*)

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2648    Accepted Submission(s): 848

Problem Description

Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

 

Input

The input contains multiple test cases.
Each test case include, first two integers n, m. (2<=n,m<=200). 
Next n lines, each line included m character.
‘Y’ express yifenfei initial position.
‘M’    express Merceki initial position.
‘#’ forbid road;
‘.’ Road.
‘@’ KCF

 

Output

For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

 

Sample Input

   
   
   
   

    
    
    
    4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    66
88
66
   
   
   
   

 

Author

yifenfei

 

Source

奋斗的年代

 

Recommend

yifenfei

题目大意：给一个迷宫，起点Y，终点M，Y和M要在@的地方相遇，求最短时间。不要求同时到达。

题目分析：给了起点和终点，很容易想到dbfs。不过这题保证有解，也可以dfs，而且这个迷宫问题很容易想到启发函数h(x) = 当前点到目标点的曼哈顿距离。所以可以IDA*。

这题先用dbfs交了一个，跑的速度还可以，不过dbfs的弱点，内存耗的太多，hdu上排100+，因为这题不要求同时到达，所以dbfs的时候不能找到解就输出，因为最先找到的并不一定是最优的。bfs先找到的解不是最优的？没错，具体的看我dbfs版本后面的测试数据。

后来看了下hdu排第一版的内存耗的少，估计是dfs做的，然后就改了个IDA*交上去，结果跑了200ms+，然后各种优化。。。还是跑不进200ms。。。

好吧，先上代码，如果我写挫了，请告诉我一声:D

dbfs:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
const int N = 205;
const int M = 10005;//205就够了。。。这数据是有多弱。。。Orz。。。
int n,m;
char map[N][N];
int flag[2][N][N];
int si,sj,ei,ej;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
struct node
{
    int x,y,time;
}ss,now;
struct que
{
    struct node t[M];
    int head,tail;
    void init()
    {
        head = tail = 0;
    }
    bool empty()
    {
        return head == tail;
    }
    void push(struct node a)
    {
        t[tail] = a;
        tail ++;
        if(tail >= M)
            tail -= M;
    }
    void pop()
    {
        head ++;
        if(head >= M)
            head -= M;
    }
    struct node top()
    {
        return t[head];
    }
}q[2];

int Min(int a,int b)
{
    return a > b?b:a;
}

int main()
{
    int i,j,k;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        getchar();
        for(i = 1;i <= n;i ++)
        {
            for(j = 1;j <= m;j ++)
            {
                map[i][j] = getchar();
                if(map[i][j] == 'Y')
                {
                    si = i;
                    sj = j;
                }
                if(map[i][j] == 'M')
                {
                    ei = i;
                    ej = j;
                }
            }
            getchar();
        }
        q[0].init();
        q[1].init();
        ss.time = 0;
        ss.x = si;
        ss.y = sj;
        memset(flag,0,sizeof(flag));
        flag[0][si][sj] = 1;
        q[0].push(ss);
        ss.x = ei;
        ss.y = ej;
        flag[1][ei][ej] = 1;
        q[1].push(ss);
        j = 0;
        int ans = 0x3fff;
        while(!q[0].empty() && !q[1].empty())
        {
            for(i = 0;i <= 1;i ++)
            {
                while(!q[i].empty() && q[i].top().time == j)
                {
                    now = q[i].top();
                    q[i].pop();
                    ss.time = now.time + 1;
                    for(k = 0;k < 4;k ++)
                    {
                        ss.x = now.x + dx[k];
                        ss.y = now.y + dy[k];
                        if(ss.x < 1 || ss.y < 1 || ss.x > n || ss.y > m || map[ss.x][ss.y] == '#')
                            continue;
                        if(!flag[i][ss.x][ss.y])
                        {
                            flag[i][ss.x][ss.y] = ss.time;
                            if(flag[1 - i][ss.x][ss.y] && map[ss.x][ss.y] == '@')
                            {
                                ans = Min(ans,flag[1 - i][ss.x][ss.y] + ss.time);
                                break;
                            }
                            q[i].push(ss);
                        }
                    }
                }
            }
            j ++;
        }
        printf("%d\n",ans * 11);
    }
    return 0;
}
//15MS	1252K
/*
5 6
Y.#...
@.....
...@#.
...##.
.....M
*/

刚很好奇的测了一下程序，发现这题的后台数据好弱的样子，循环队列开205就够了。。。。。结果内存也是大减，瞬间跑第二版了。。。

IDA*:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
const int N = 205;
const int M = 1000005;
int n,m;
int ans;
char map[N][N];
int flag[2][N][N];//这题判不判重都行
int si,sj,ei,ej;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
bool ok;

int Abs(int a)
{
    return a > 0?a:-a;
}

void Dfs(int curi,int curj,int dp)
{
    if(ok)
        return;
    if(dp + Abs(curi - ei) + Abs(curj - ej) > ans)
        return;
    if(map[curi][curj] == 'M')
    {
        ok = true;
        return;
    }
    int i;
    //flag[1][curi][curj] = 1;
    for(i = 0;i < 4;i ++)
    {
        int tx = curi + dx[i];
        int ty = curj + dy[i];
        if(tx < 1 || ty < 1 || tx > n || ty > m || map[tx][ty] == '#')// || flag[1][tx][ty])
            continue;
        Dfs(tx,ty,dp + 1);
    }
    //flag[1][curi][curj] = 0;
}

void dfs(int curi,int curj,int dp)
{
    if(ok)
        return;
    if(dp + Abs(curi - ei) + Abs(curj - ej) > ans)
        return;
    //flag[0][curi][curj] = 1;
    if(map[curi][curj] == '@')
    {
        Dfs(curi,curj,dp);
    }
    int i;
    for(i = 0;i < 4;i ++)
    {
        int tx = curi + dx[i];
        int ty = curj + dy[i];
        if(tx < 1 || ty < 1 || tx > n || ty > m || map[tx][ty] == '#')// || flag[0][tx][ty])
            continue;
            dfs(tx,ty,dp + 1);
    }
    //flag[0][curi][curj] = 0;
}

int main()
{
    int i,j,k;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        //memset(flag,0,sizeof(flag));
        getchar();
        for(i = 1;i <= n;i ++)
        {
            for(j = 1;j <= m;j ++)
            {
                map[i][j] = getchar();
                if(map[i][j] == 'Y')
                {
                    si = i;
                    sj = j;
                }
                if(map[i][j] == 'M')
                {
                    ei = i;
                    ej = j;
                }
//                if(map[i][j] == '#')
//                    flag[0][i][j] = flag[1][i][j] = 1;
            }
            getchar();
        }
        ans = Abs(si - ei) + Abs(sj - ej);//迭代初始深度不必从1开始
        ok = false;
        while(!ok)
        {
            dfs(si,sj,0);
            ans ++;
        }
        printf("%d\n",ans * 11 - 11);
    }
    return 0;
}
//265MS	440K
//203MS	604K
//为什么会跑这么慢。。。