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poj 1979 dfs水题

时间:2015-05-13 14:45:49      阅读:107      评论:0      收藏:0      [点我收藏+]

标签:poj 1979   dfs水题   

// 练练水题,夯实基础吧
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#define ceil(a,b) (((a)+(b)-1)/(b))
#define endl '\n'
#define gcd __gcd
#define highBit(x) (1ULL<<(63-__builtin_clzll(x)))
#define popCount __builtin_popcountll
typedef long long ll;
using namespace std;
const int MOD = 1000000007;
const long double PI = acos(-1.L);

template<class T> inline T lcm(const T& a, const T& b) { return a/gcd(a, b)*b; }
template<class T> inline T lowBit(const T& x) { return x&-x; }
template<class T> inline T maximize(T& a, const T& b) { return a=a<b?b:a; }
template<class T> inline T minimize(T& a, const T& b) { return a=a<b?a:b; }

const int maxn = 32;
char g[maxn][maxn];
int n,m;
bool vis[maxn][maxn];
int dx[4] = {-1,0,1,0};
int dy[4] = {0,1,0,-1};
int cnt ;
void init(){
	for (int i=0;i<n;i++)
		scanf("%s",g[i]);
	cnt = 0;
	memset(vis,0,sizeof(vis));
}

void dfs(int x,int y){
//	g[x][y] = '#';
	vis[x][y] = 1;
	//cnt = max(num,cnt);
	cnt++;
	for (int i=0;i<4;i++){
		int tx = x + dx[i];
		int ty = y + dy[i];
		if (tx>=n||tx<0||ty>=m||ty<0)
			continue;
		if (g[tx][ty]=='#')
			continue;
		if (vis[tx][ty])
			continue;
		dfs(tx,ty);
	}
}

void solve(){
	for (int i=0;i<n;i++)
		for (int j=0;j<m;j++)
			if (g[i][j] == '@')
				dfs(i,j);
	printf("%d\n",cnt);
}

int main() {
	//freopen("G:\\Code\\1.txt","r",stdin);
	while(scanf("%d%d",&m,&n)!=EOF){
		if (n==0&&m==0)
			break;
		init();
		solve();
	}
	return 0;
}

poj 1979 dfs水题

标签:poj 1979   dfs水题   

原文地址:http://blog.csdn.net/timelimite/article/details/45692923

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