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POJ 2386 Lake Counting (水题,DFS)

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;}

 

POJ 2386 Lake Counting (水题,DFS)