搜索问题合集

1.POJ 1979

Language:
Red and Black
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 46232 Accepted: 24899
Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
The end of the input is indicated by a line consisting of two zeros.
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

深度优先搜索问题，对这个人四个方向进行Dfs搜索，当寻找到‘.’，递归地调用Dfs函数，每搜索一次将后将搜索过的位置标记。

#include <iostream>
#include <vector>
#include <string>
#define MaxSize 21
using namespace std;
vector<string> matrix(MaxSize);
int W, H;

int Dfs(int x,int y) {
    int num = 1;
    for(int i=-1;i<=1;i++)
        for (int j = -1; j <= 1; j++) {
            if ((i + j == 1 || i + j == -1)&&i+x<W&&i+x>=0&&j+y<H&&j+y>=0) {
                if (matrix[y + j][x + i]=='.') {
                    matrix[y + j][x + i] = '#';
                    num += Dfs(i + x, j + y);
                }
            }
        }
    return num;
}

int main()
{
    vector<int> output;
    while (cin >> W >> H) {
        if (W == 0 && H == 0) 
            break;
        int x=0, y=0;
        bool hasFind = false;
        
        for (int i = 0; i < H; i++) {
            cin >> matrix[i];
            if (!hasFind) {
                int index = matrix[i].find('@');
                if (index != -1) {
                    x = index;
                    y = i;
                    hasFind = true;
                }
            }
        }
        output.push_back(Dfs(x, y));
    }
    for (int i = 0; i < output.size(); i++) {
        cout << output[i] << endl;
    }
    return 0;
}