POJ 2488 A Knight's Journey 题解

Porridge — Sun, 08 Sep 2013 12:31:00 GMT

A Knight's Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 26210 Accepted: 8950
Description

Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.
Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .
Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.
Sample Input

3
1 1
2 3
4 3
Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
Source

TUD Programming Contest 2005, Darmstadt, Germany

因为题目要求 print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line.即输出字典序最小的一组答案

字典序最小就是按位比较ASCII码，相等的比较下一位，知道不相等的为止或者长度比另一个长比如 "abc" < "bbc" "abcd" > "abc" 因为要是有路径的话长度都是一样所以不需要考虑路径长度的问题。

直接考虑每个路径，在搜索的时候只要优先往A副1的方向走，这样获得的第一条路径就是字典序最小的路径了。按照图上的路径。有的题解是转过来的，其实只要写程序的时候注意一下就行了。

这个题目我WA了N多次的原因是Scenario自己打的没有复制，然后输错了。太二了

POJ2488.cpp
#include
#include
#include
using namespace std;

bool vis[26][26];
char anx[800],any[800];
char ax[800],ay[800];
int ind = 0;
const int mx[] = {-2,-2,-1,-1,+1,+1,+2,+2};
const int my[] = {-1,+1,-2,+2,-2,+2,-1,+1};
int p,q;
int get = 0;
void push(int x,int y)
{
    anx[ind] = x;
    any[ind] = y;
    ind++;
}
void pop()
{
    ind--;
}
bool check(int x,int y)
{
    return  (((x >= 0 && x < p)&&(y >= 0 && y < q))&&!vis[x][y]);
}
void init()
{
    get = 0;
    memset(vis,0,sizeof(vis));
    ind = 0;
}
bool betteranswer()
{
    int bt = 0;
    for(int i = 0;i < p*q;i++)
    {
        if (anx[i] < ax[i])
        {
            bt = 1;
            break;
        }
        else if (anx[i] == ax[i] && any[i] < ay[i])
        {
            bt = 1;
            break;
        }
        else if (anx[i] == ax[i] && any[i] == ay[i])
            continue;
        else
            break;
    }
    return bt;
}
void dfs(int i,int x,int y)
{
    push(x,y);
    vis[x][y] = 1;
    if (i == p*q-1)
    {
        if (get&& betteranswer())
        {
            memcpy(ax,anx,sizeof(ax));
            memcpy(ay,any,sizeof(ay));
            return;
        }
        else if (!get)
        {
            get = 1;
            memcpy(ax,anx,sizeof(ax));
            memcpy(ay,any,sizeof(ay));
        }
        pop();
        vis[x][y] = 0;
        return;
    }
    for(int j = 0;j < 8;j++)
    {
        int tx = x + mx[j];
        int ty = y + my[j];
        if (check(tx,ty))
            dfs(i+1,tx,ty);
    }
    pop();
    vis[x][y] = 0;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    while(scanf("%d",&T)!=EOF)
    {
        int f = 0;
        for(int i = 1;i <= T;i++)
        {
            scanf("%d%d",&q,&p);
            init();
            for(int j = 0;j < p && !get ;j++)
            {
                for(int k = 0;k < q && !get;k++)
                {
                    dfs(0,j,k);
                }
            }
            if (!f)
            {
                f = 1;
            }
            else
            {
                printf("\n");
            }
            printf("Scenario #%d:\n",i);
            if (get)
            {
                for(int i = 0;i < p*q;i++)
                {
                    printf("%c%d",ax[i]+65,ay[i]+1);
                }
                printf("\n");
            }
            else
            {
                printf("impossible\n");
            }
        }
    }
    return 0;
}

Porridge 2013-09-08 20:31 发表评论

POJ 1321 棋盘问题题解 DFS

Porridge — Sat, 07 Sep 2013 11:18:00 GMT

棋盘问题
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
Description
在一个给定形状的棋盘（形状可能是不规则的）上面摆放棋子，棋子没有区别。要求摆放时任意的两个棋子不能放在棋盘中的同一行或者同一列，请编程求解对于给定形状和大小的棋盘，摆放k个棋子的所有可行的摆放方案C。
Input
输入含有多组测试数据。
每组数据的第一行是两个正整数，n k，用一个空格隔开，表示了将在一个n*n的矩阵内描述棋盘，以及摆放棋子的数目。 n <= 8 , k <= n
当为-1 -1时表示输入结束。
随后的n行描述了棋盘的形状：每行有n个字符，其中 # 表示棋盘区域， . 表示空白区域（数据保证不出现多余的空白行或者空白列）。
Output
对于每一组数据，给出一行输出，输出摆放的方案数目C （数据保证C<2^31）。
Sample Input
2 1
#.
.#
4 4
...#
..#.
.#..
#...
-1 -1
Sample Output
2
1

简单的暴力DFS
从棋盘的第一行开始。往下搜知道到达规定的步数或者出界了。其中加DFS中的最后一句是因为这一行可以不占。然后往下一行。

POJ1321.cpp

Porridge 2013-09-07 19:18 发表评论

C++博客-RePorridge-随笔分类-POJ题解

POJ 2488 A Knight's Journey 题解

POJ 1321 棋盘问题 题解 DFS

POJ 1321 棋盘问题题解 DFS