Arthur and Walls CodeForces - 525D (bfs)

大意: 给定格点图, 每个'.'的连通块会扩散为矩形, 求最后图案.

一开始想得是直接并查集合并然后差分, 但实际上是不对的, 这个数据就可以hack掉.

3 3

**.

.**

...

正解是bfs, 一个点被扩散当且仅当它所在的某个2*2块中只有它为'*'.

#include <iostream>

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+10;

#else

const int N = 111;

#endif

int n, m;

char s[N][N];

int dx[]={-1,-1,-1,0,1,1,1,0};

int dy[]={-1,0,1,1,1,0,-1,-1};

queue<pii> q;

int chk(int x, int y) {

	if (x<1||x>n||y<1||y>m||s[x][y]=='.') return 0;

	if (s[x-1][y]=='.'&&s[x-1][y-1]=='.'&&s[x][y-1]=='.') return 1;

	if (s[x-1][y]=='.'&&s[x-1][y+1]=='.'&&s[x][y+1]=='.') return 1;

	if (s[x][y-1]=='.'&&s[x+1][y-1]=='.'&&s[x+1][y]=='.') return 1;

	if (s[x][y+1]=='.'&&s[x+1][y]=='.'&&s[x+1][y+1]=='.') return 1;

	return 0;

}

int main() {

	scanf("%d%d", &n, &m);

	REP(i,1,n) scanf("%s", s[i]+1);

	REP(i,1,n) REP(j,1,m) if (chk(i,j)) q.push(pii(i,j));

	while (q.size()) {

		auto t = q.front(); q.pop();

		if (!chk(t.x,t.y)) continue;

		s[t.x][t.y] = '.';

		REP(i,0,7) {

			int x=t.x+dx[i],y=t.y+dy[i];

			if (chk(x,y)) q.push(pii(x,y));

		}

	}

	REP(i,1,n) puts(s[i]+1);

}