2815:城堡问题 输入处理，连通块

Posted 2020-11-02 COLORFUL

http://bailian.openjudge.cn/practice/2815?lang=en_US

题意：给你一个地图（二进制处理） 问有几个连通块，与最大连通块的面积。

题解：处理输入时，用四个if位运算判断。连通块模板处理。

#define _CRT_SECURE_NO_WARNINGS
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<cstdio>
#include<string>
#include<stack>
#include<ctime>
#include<list>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<sstream>
#include<iostream>
#include<functional>
#include<algorithm>
#include<memory.h>
//#define INF 0x3f3f3f3f
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
#define rep(i,t,n)  for(int i =(t);i<=(n);++i)
#define per(i,n,t)  for(int i =(n);i>=(t);--i)
#define mp make_pair
#define pb push_back
#define mmm(a,b) memset(a,b,sizeof(a))
//std::ios::sync_with_stdio(false);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
void smain();
#define ONLINE_JUDGE
int main() {
    ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    long _begin_time = clock();
#endif
    smain();
#ifndef ONLINE_JUDGE
    long _end_time = clock();
    printf("time = %ld ms.", _end_time - _begin_time);
#endif
    return 0;
}
const int maxn = 1e4 + 100;
const ll mod = 1e5 + 7;
const ll INF = (100000)*(200000ll) + 1000;

int n, m;
int rooms[55][55];
int color[55][55];
int roomNum, mx;
int area;
void dfs(int i,int k) {
    if (color[i][k])return;
    area++;
    color[i][k] = roomNum;
    int x = rooms[i][k];

    if ((x&1) == 0)dfs(i, k - 1);
    
    if ((x & 2) == 0)dfs(i-1, k);
    
    if ((x & 4) == 0)dfs(i, k + 1);
    
    if ((x & 8)==0)dfs(i+1, k);
    

}
void Run() {

}

void smain() {
    cin >> n >> m;
    rep(i,1,n)
        rep(j, 1, m) {
        cin>> rooms[i][j];
}
    mmm(color, 0);
    rep(i, 1, n)
        rep(j, 1, m) {
        if (!color[i][j]) {
            roomNum++;
            area = 0;
            dfs(i, j);
            mx = max(area, mx);
        }
    }
    cout << roomNum << endl;
    cout << mx << endl;
}