HDU1312 Red and Black 题解

一条递归搜索法题目，使用递归搜索法，但是实际不用重复计算方格。

思路是：

1 每搜索一个方格就改变当前方格的值为 ‘*’，或者任何其他非‘.‘的值，代表该方格已经走过了

2 递归的时候不回复这个方格的值，就实际上不用重复搜索这个方格了，故此不用回溯

#include <stdio.h>
#include <iostream>
#include <string>
#include <vector>
using namespace std;
int R, C, blacks;
vector<string> board;

inline bool isLegal(int r, int c)
{
    return r>=0 && c>=0 && r<R && c<C && board[r][c] == '.';
}

void getBlacks(int r, int c)
{
    blacks++;
    board[r][c] = '*';
    if (isLegal(r-1, c)) getBlacks(r-1, c);
    if (isLegal(r+1, c)) getBlacks(r+1, c);
    if (isLegal(r, c-1)) getBlacks(r, c-1);
    if (isLegal(r, c+1)) getBlacks(r, c+1);
}

int main()
{
    string s;
    while (cin>>C>>R && C)
    {
        board.clear();
        for (int i = 0; i < R; i++)
        {
            cin>>s;
            board.push_back(s);
        }
        blacks = 0;
        for (unsigned i = 0; i < board.size(); i++)
        {
            for (unsigned j = 0; j < board[0].size(); j++)
            {
                if ('@' == board[i][j])
                {
                    getBlacks((int)i, (int)j);
                    goto out;
                }
            }
        }
out:;
        printf("%d\n", blacks);
    }
    return 0;
}

HDU1312 Red and Black 题解