POJ 2386 Lake Counting (水题，DFS)

Posted 2020-08-05 dwtfukgv

题意：给定一个n*m的矩阵，让你判断有多少个连通块。

析：用DFS搜一下即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <list>
#include <sstream>
#define frer freopen("in.txt", "r", stdin)
#define frew freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e2 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn][maxn];
int vis[maxn][maxn];

void dfs(int r, int c){
    vis[r][c] = 1;
    for(int i = 0; i < 8; ++i){
        int x = r + dr[i];
        int y = c + dc[i];
        if(is_in(x, y) && !vis[x][y] && s[x][y] == ‘W‘)  dfs(x, y);
    }
}
int main(){
    while(scanf("%d %d", &n, &m) == 2){
        for(int i = 0; i < n; ++i)  scanf("%s", s[i]);
        memset(vis, 0, sizeof vis);
        int ans = 0;
        for(int i = 0; i < n; ++i)
            for(int j = 0; j < m; ++j)
                if(!vis[i][j] && s[i][j] == ‘W‘)  dfs(i, j), ++ans;
        printf("%d\n", ans);
    }
    return 0;
}