深度优先搜索

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深度优先搜索

基本原理:

从一个状态开始不断转移,直至无法转移,然后退回前一步的状态,继续转移到其他状态,知道求解

基本思想:递归

例一:

给定整数a1,a2,a3......an,判断是否可以从中选出若干数,使们的和恰好为K;

#include<iostream>
using namespace std;
int a[100];
bool dfs(int i, int sum, int n, int k)
{
    if(i == n )
        return sum == k;
    if(dfs(i + 1, sum , n, k))
        return true;
    if(dfs(i + 1, sum + a[i], n, k))
        return true;
    return false;
}

int main()
{
    int n, k;
    cin >> n >> k;
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    bool t = dfs(0, 0, n, k);
    if(t == true)
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

例二:

Due to recent rains, water has pooled in various places in Farmer John‘s field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water (‘W‘) or dry land (‘.‘). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John‘s field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John‘s field. Each character is either ‘W‘ or ‘.‘. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John‘s field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

3

思路:运用深度优先搜索的方法,首先确定一个w,将其相邻的w都变成‘.’,这样,每调用一次dfs函数,就会消去一处水洼,dfs函数的执行次数就是水洼数;

#include<iostream>
using namespace std;
int N, M;
char field[100][100];

void dfs(int x, int y)
{
    field[x][y] = '.'; 
    //两层循环,依次遍历与该点相邻的点
    for(int dx = -1; dx <= 1; dx++){
        for(int dy = -1; dy <= 1; dy++){
            int nx = x + dx, ny = y + dy;
            if(nx >= 0 && nx < N && ny >= 0 && ny < M && field[nx][ny] == 'w')
                dfs(nx, ny);
        }
    }
}

int main()
{
    cin >> N >> M;
    for(int i = 0; i < N; i++){
        for(int j = 0; j < M; j++){
            cin >> field[i][j];
        }
    }
    int sum = 0;
    for(int i = 0; i < N; i++){
        for(int j = 0; j < M; j++){
            if(field[i][j] == 'w')
            {
                dfs(i,j);
                sum++;
            }
            
        }
    }
    cout << sum << endl;
    return 0;
}

深度优先搜索

标签：end   int   后退   ret   lan   line   无法   repr   优先   

原文地址：https://www.cnblogs.com/zhulmz/p/12205193.html