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;}