Arthur and Walls CodeForces - 525D (bfs)

最新推荐文章于 2020-10-19 20:43:00 发布

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最新推荐文章于 2020-10-19 20:43:00 发布

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原文链接：http://www.cnblogs.com/uid001/p/10779423.html

本文探讨了格点图中特定扩散规则的应用，通过BFS算法实现连通块的矩形扩散，解决复杂图案生成问题。文章分析了并查集方法的局限性，并详细解释了BFS算法的正确应用。

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);
}

转载于:https://www.cnblogs.com/uid001/p/10779423.html