poj 1154 LETTERS

最新推荐文章于 2025-03-30 16:45:00 发布

chaoyueziji123

最新推荐文章于 2025-03-30 16:45:00 发布

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分类专栏：深搜

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本文介绍了一个基于深搜的游戏路径探索算法。游戏在一个由字母组成的矩形棋盘上进行，目标是尽可能多地移动游戏棋子，但不能两次进入同一个字母的位置。文章提供了完整的C++代码实现，并解释了如何通过深度优先搜索寻找最长路径。

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LETTERS

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6009		Accepted: 2847

Description

A single-player game is played on a rectangular board divided in R rows and C columns. There is a single uppercase letter (A-Z) written in every position in the board.
Before the begging of the game there is a figure in the upper-left corner of the board (first row, first column). In every move, a player can move the figure to the one of the adjacent positions (up, down,left or right). Only constraint is that a figure cannot visit a position marked with the same letter twice.
The goal of the game is to play as many moves as possible.
Write a program that will calculate the maximal number of positions in the board the figure can visit in a single game.

Input

The first line of the input contains two integers R and C, separated by a single blank character, 1 <= R, S <= 20.
The following R lines contain S characters each. Each line represents one row in the board.

Output

The first and only line of the output should contain the maximal number of position in the board the figure can visit.

Sample Input

3 6
HFDFFB
AJHGDH
DGAGEH

Sample Output

分析：简单的深搜从（0,0）开始进行搜索，然后找到最长的路。需要将每一个字母进行标记。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char map[101][101];
bool vis[101];
int dir[4][2]={1,0,0,1,-1,0,0,-1};
int ans,n,m;
int check(int x,int y)
{
    if(x>=0&&x<n&&y>=0&&y<m&&!vis[map[x][y]-'A'])
    return 1;
    return 0;
}
void dfs(int dep,int x,int y)
{
    if(dep>ans)
    {
        ans=dep;
    }
    for(int i=0;i<4;i++)
    {
        int tx=x+dir[i][0];
        int ty=y+dir[i][1];
        if(check(tx,ty))
        {
            vis[map[tx][ty]-'A']=true;
            dfs(dep+1,tx,ty);
            vis[map[tx][ty]-'A']=false;
        }
    }
}
int main()
{
    while(scanf("%d %d",&n,&m)==2)
    {
        ans=0;
        memset(vis,false,sizeof(vis));
        for(int i=0;i<n;i++)
        {
           for(int j=0;j<m;j++)
           {
               cin>>map[i][j];
               map[i][j]-='A';
           }
        }
        vis[map[0][0]-'A']=true;
        dfs(1,0,0);
        cout<<ans<<endl;
    }
    return 0;
}