BFS，DFS 典型题目总结

wkh2021

已于 2023-08-09 08:39:22 修改

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文章标签：深度优先宽度优先算法图论

于 2022-02-25 14:39:52 首次发布

本文链接：https://blog.youkuaiyun.com/wkh2021/article/details/123132313

这篇博客介绍了两种图论算法——BFS（广度优先搜索）和DFS（深度优先搜索）在解决最短路径问题和计算连通块个数及最大面积问题上的应用。通过示例题目‘仙岛求药’和‘最大岛屿’，展示了如何使用这两种算法来寻找迷宫中的最短路径以及在二维矩阵中找到最大岛屿的面积。BFS适用于寻找最短路径，而DFS则用于查找连通组件并计算其数量和面积。

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1. 最短路径+判断无解：

例题： ybt 1251：仙岛求药

BFS：

#include <bits/stdc++.h>
using namespace std;
 
const int N = 105;
 
constexpr int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
 
struct node { int x, y, step; } S, E;
 
int m, n;
char a[N][N];
bool vis[N][N];
 
inline bool input() {
	scanf("%d%d", &m, &n);
	if (m == 0 && n == 0) return false;
	for (int i = 1; i <= m; i++) {
		scanf("%s", a[i] + 1);
		for (int j = 1; j <= n; j++) {
			if(a[i][j] == '@') S = {i, j};
			if(a[i][j] == '*') E = {i, j};
		}
	}
	memset(vis, false, sizeof(vis));
	return true;
}
 
inline void bfs() {
	auto isValid = [](int x, int y) {
		return (1 <= x && x <= m && 1 <= y && y <= n && a[x][y] != '#' && !vis[x][y]);
	};
	queue<node> q; q.push(S);
 
	while (!q.empty()) {
		node u = q.front(); q.pop();
 
		for(int i = 0; i < 4; i++) {
			int _x = u.x + dx[i], _y = u.y + dy[i];
 
			if (!isValid(_x, _y)) continue;
			
			if (_x == E.x && _y == E.y) {
				printf("%d\n", u.step + 1);
				return;
			}
			q.push({_x, _y, u.step + 1});
			vis[_x][_y] = true;
		}
	}
	printf("-1\n");
}
 
signed main() {
	while (input()) bfs();
	return 0;
}

DFS：

#include <bits/stdc++.h>
using namespace std;
 
const int N = 105;
 
constexpr int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
 
int m, n, step[N][N];
char a[N][N];
bool hasAns;
 
int st_x, st_y, ed_x, ed_y;
 
inline bool input() {
	scanf("%d%d", &m, &n);
	if (m == 0 && n == 0) return false;
	for (int i = 1; i <= m; i++) {
		scanf("%s", a[i] + 1);
		for (int j = 1; j <= n; j++) {
			if(a[i][j] == '@') st_x = i, st_y = j;
			if(a[i][j] == '*') ed_x = i, ed_y = j;
		}
	}
	memset(step, 0x3f, sizeof(step)); step[st_x][st_y] = 0;
	return true;
}
 
inline bool isValid(int x, int y) {
	return (1 <= x && x <= m && 1 <= y && y <= n && a[x][y] != '#');
};
 
inline void dfs(int x, int y) {
	if (a[x][y] == '*') {
		hasAns = true;
		return;
	}
	for(int i = 0; i < 4; i++) {
		int _x = x + dx[i], _y = y + dy[i];
 
		if (!isValid(_x, _y)) continue;
		if (step[_x][_y] <= step[x][y] + 1) continue;
 
		step[_x][_y] = step[x][y] + 1;
		dfs(_x, _y);
	}
}
 
signed main() {
	while (input()) {
		hasAns = false;
		dfs(st_x, st_y);
		if (hasAns) printf("%d\n", step[ed_x][ed_y]);
		else printf("-1\n");
	}
	return 0;
}

2. 连通块个数+最大面积：

例题：计蒜客 T1405 最大岛屿

BFS：

#include <bits/stdc++.h>
using namespace std;

const int N = 505;

constexpr int dx[]= {0, 0, 1, -1, 1, -1, 1, -1};
constexpr int dy[]= {1, -1, 0, 0, 1, -1, -1, 1};
 
struct node { int x, y; };

int m, n, t;
char a[N][N];
bool vis[N][N];

int cnt, ans, now;
 
inline void bfs(int x,int y) {
    auto isValid = [](int x, int y) {
        return (1 <= x && x <= m && 1 <= y && y <= n && a[x][y] != '0' && !vis[x][y]);
    };
    
	queue<node> q; q.push({x, y});
 
	while (!q.empty()) {
		node u = q.front(); q.pop();
 
		for (int i = 0; i < 8; i++) {
            int _x = u.x + dx[i], _y = u.y + dy[i];
			if (!isValid(_x, _y)) continue;
            vis[_x][_y] = true;
			now++;
			q.push({_x, _y});
		}
	}
}
 
inline void input() {
	scanf("%d%d%d", &m, &n, &t);
	for (int i = 1; i <= m; i++)
		for (int j = 1; j <= n; j++) {
			do {
				a[i][j] = getchar();
			} while (a[i][j] != '0' && a[i][j] != '1');
		}
}
 
inline void solve() {
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			if (a[i][j] == '0' || vis[i][j]) continue;
            vis[i][j] = true;
			now = 1;
			bfs(i, j);
            ans = max(ans, now);
			cnt++;
		}
	}
	printf("%d %d\n", cnt, ans * t);
}
 
signed main() {
	input();
	solve();
	return 0;
}

DFS：

#include <bits/stdc++.h>
using namespace std;

const int N = 505;

constexpr int dx[]= {0, 0, 1, -1, 1, -1, 1, -1};
constexpr int dy[]= {1, -1, 0, 0, 1, -1, -1, 1};
 
int m, n, t;
char a[N][N];
bool vis[N][N];
 
int cnt, ans, now;
 
inline void input() {
	scanf("%d%d%d", &m, &n, &t);
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			do {
				a[i][j] = getchar();
			} while (a[i][j] != '0' && a[i][j] != '1');
		}
	}
}
 
inline bool isValid (int x, int y) {
    return (1 <= x && x <= m && 1 <= y && y <= n && a[x][y] != '0' && !vis[x][y]);
}

inline void dfs(int x,int y) {
	for (int i = 0; i < 8; i++) {
        int _x = x + dx[i], _y = y + dy[i];
		if (!isValid(_x, _y)) continue;
        vis[_x][_y] = true;
        now++;
		dfs(_x,_y);
	}
}
 
inline void solve() {
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			if (a[i][j] == '0' || vis[i][j]) continue;
            vis[i][j] = true;
			now = 1;
			dfs(i, j);
            ans = max(ans, now);
			cnt++;
		}
	}
	printf("%d %d\n", cnt, ans * t);
}
 
signed main() {
	input();
	solve();
	return 0;
}