hdu 1240 Asteroids!

最新推荐文章于 2021-08-21 23:31:09 发布

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本文深入探讨了在三维空间内利用BFS和DFS算法进行路径搜索的方法，并通过实例代码展示了如何求解从出发点到目标点所需的最短步数。同时，对比了两种算法在解决此类问题时的效率和代码简洁性。

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Asteroids!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4530    Accepted Submission(s): 2920


Problem Description
You're in space.
You want to get home.
There are asteroids.
You don't want to hit them.


Input
Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.

A single data set has 5 components:

Start line - A single line, "START N", where 1 <= N <= 10.

Slice list - A series of N slices. Each slice is an N x N matrix representing a horizontal slice through the asteroid field. Each position in the matrix will be one of two values:

'O' - (the letter "oh") Empty space

'X' - (upper-case) Asteroid present

Starting Position - A single line, "A B C", denoting the <A,B,C> coordinates of your craft's starting position. The coordinate values will be integers separated by individual spaces.

Target Position - A single line, "D E F", denoting the <D,E,F> coordinates of your target's position. The coordinate values will be integers separated by individual spaces.

End line - A single line, "END"

The origin of the coordinate system is <0,0,0>. Therefore, each component of each coordinate vector will be an integer between 0 and N-1, inclusive.

The first coordinate in a set indicates the column. Left column = 0.

The second coordinate in a set indicates the row. Top row = 0.

The third coordinate in a set indicates the slice. First slice = 0.

Both the Starting Position and the Target Position will be in empty space.



Output
For each data set, there will be exactly one output set, and there will be no blank lines separating output sets.

A single output set consists of a single line. If a route exists, the line will be in the format "X Y", where X is the same as N from the corresponding input data set and Y is the least number of moves necessary to get your ship from the starting position to the target position. If there is no route from the starting position to the target position, the line will be "NO ROUTE" instead.

A move can only be in one of the six basic directions: up, down, left, right, forward, back. Phrased more precisely, a move will either increment or decrement a single component of your current position vector by 1.



Sample Input
START 1
O
0 0 0
0 0 0
END
START 3
XXX
XXX
XXX
OOO
OOO
OOO
XXX
XXX
XXX
0 0 1
2 2 1
END
START 5
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
XXXXX
XXXXX
XXXXX
XXXXX
XXXXX
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
0 0 0
4 4 4
END


Sample Output
1 0
3 4
NO ROUTE


Source
South Central USA 2001

题目大意：在三维空间里搜索从出发点到中止点需要的最短步数
采用了两种方法，BFS和DFS 我个人更喜欢使用BFS，但好几次做
搜索题对比之后发现通常情况下DFS比BFS写起来代码更简洁一些，
思路只要完整，优化剪枝之后效率比BFS更高。此题DFS 0ms通过
看来以后要多使用DFS了。

BFS代码实现：

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
#include<string.h>
#include<map>
#include<queue>
#include<stack>
#include<list>
#include<cctype>
#include<fstream>
#include<sstream>
#include<iomanip>
#include<set>
#include<vector>
#include<cstdlib>
#include<time.h>
using namespace std;

struct node{
    char c;
    int x, y, z;
    bool book;
    int dis;
    friend bool operator ==(node a,node b)
    {
        return a.x == b.x&&a.y == b.y&&a.z == b.z;
    }
} m[12][12][12];
int fx[6][3] = { { 1, 0, 0 }, { 0, 1, 0 }, { 0, 0, 1 }, { -1, 0, 0 }, { 0, -1, 0 }, { 0, 0, -1 } };
node start, target;
int n;
int good(int a, int b, int c)
{
    if (m[a][b][c].c=='O'&&m[a][b][c].book == 0 && a >= 0 && a < n&&b >= 0 && b < n&&c >= 0 && c < n)
        return 1;
    else if (a == target.x&&b == target.y&&c == target.z)
        return 2;
    else
        return 3;
}
int bfs()
{
    queue<node> duilie;
    duilie.push(start);
    m[start.x][start.y][start.z].book = true;
    while (!duilie.empty())
    {
        node t = duilie.front();
        duilie.pop();
        for (int i = 0; i < 6; i++)
        {
            int a, b, c;
            a = t.x + fx[i][0];
            b = t.y + fx[i][1]; 
            c = t.z + fx[i][2];
            if (good(a, b, c)<=2)
            {
                m[a][b][c].book = 1;
                m[a][b][c].dis = m[t.x][t.y][t.z].dis+1;
                if (good(a, b, c) == 2)
                    return m[a][b][c].dis;
                duilie.push(m[a][b][c]);
            }
        }
    }
    return -1;
}


int main()
{
    string sta;
    while (cin >> sta >> n)
    {
        for (int k = 0; k < n; k++)
        {
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    cin >> m[i][j][k].c;
                    m[i][j][k].x = i;
                    m[i][j][k].y = j;
                    m[i][j][k].z = k;
                    m[i][j][k].book = false;
                    m[i][j][k].dis = 0;
                }
            }
        }
        cin >> start.x >> start.y >> start.z;
        cin >> target.x >> target.y >> target.z;
        cin >> sta;
        int s = bfs();
        if (start == target)
            cout << n << ' ' << 0 << endl;
        else
        {
            if (s == -1)
                cout << "NO ROUTE" << endl;
            else
                cout << n << ' ' << s << endl;
        }
    }
    return 0;
}

DFS代码实现

#include<iostream>
#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstring>
#include<string.h>
#include<map>
#include<queue>
#include<stack>
#include<list>
#include<cctype>
#include<fstream>
#include<sstream>
#include<iomanip>
#include<set>
#include<vector>
#include<cstdlib>
#include<time.h>
using namespace std;

struct node{
    char c;
    int x, y, z;
    int dis;
    friend bool operator ==(node a, node b)
    {
        return a.x == b.x&&a.y == b.y&&a.z == b.z;
    }
} m[12][12][12];
int fx[6][3] = { { 1, 0, 0 }, { 0, 1, 0 }, { 0, 0, 1 }, { -1, 0, 0 }, { 0, -1, 0 }, { 0, 0, -1 } };
node start, target;
int n;
int good(int a, int b, int c)
{
    if (a == target.x&&b == target.y&&c == target.z)
        return 2;
    else if (m[a][b][c].c == 'O'&& a >= 0 && a < n&&b >= 0 && b < n&&c >= 0 && c < n)
        return 1;
    else if (m[a][b][c].c == 'X')
        return 3;
    else
        return 4;
}
int maxs = INT_MAX;
void dfs(int a, int b, int c, int cnt)
{
    if (good(a, b, c) == 2)
    {
        if (cnt < maxs)
            maxs = cnt;
        return;
    }
    if (a < 0 || b < 0 || c<0 || a >= n || b >= n || c >= n)
        return;
    if (m[a][b][c].c == 'X')
        return;
    if (cnt>m[a][b][c].dis)
        return;
    else
        m[a][b][c].dis = cnt;
    for (int i = 0; i < 6; i++)
    {
        int aa, bb, cc;
        aa = a + fx[i][0];
        bb = b + fx[i][1];
        cc = c + fx[i][2];
        dfs(aa, bb, cc, cnt + 1);
    }
}


int main()
{
    string sta;
    while (cin >> sta >> n)
    {
        for (int k = 0; k < n; k++)
        {
            for (int i = 0; i < n; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    cin >> m[i][j][k].c;
                    m[i][j][k].x = i;
                    m[i][j][k].y = j;
                    m[i][j][k].z = k;
                    m[i][j][k].dis = INT_MAX;
                }
            }
        }
        cin >> start.x >> start.y >> start.z;
        cin >> target.x >> target.y >> target.z;
        cin >> sta;
        maxs = INT_MAX;
        dfs(start.x, start.y, start.z, 0);
        if (start == target)
            cout << n << ' ' << 0 << endl;
        else
        {
            if (maxs == INT_MAX)
                cout << "NO ROUTE" << endl;
            else
                cout << n << ' ' << maxs << endl;
        }
    }
    return 0;
}