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ABC-383-C - Humidifier 3 （多源广搜）

Posted on Edited on

思路讲解

多源广搜经典套路，如果该点到达时间没有别的点早，就可以不用走了

然后把所有点加入q，实现多源广搜

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FOR(i,1,N){
	FOR(j,1,M){
		if(maze[i][j]!='H') continue;
		q.push({i,j,0});
		vis[i][j]=0;
	}
}
bfs();

AC代码

https://atcoder.jp/contests/abc383/submissions/62366428

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// Problem: C - Humidifier 3
// Contest: AtCoder - Daiwa Securities Co. Ltd. Programming Contest 2024（AtCoder Beginner Contest 383）
// URL: https://atcoder.jp/contests/abc383/tasks/abc383_c
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define ROF(i, a, b) for (int i = a; i >= b; --i)
#define all(x) x.begin(),x.end()
#define debug(x) cerr<<x<<' '
#define CLR(i,a) memset(i,a,sizeof(i))

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr;
typedef double DB;
const ll MAXN=static_cast<ll>(1e3)+10,MAXval=static_cast<ll>(5e18)+3;

ll N,M,D;
char maze[MAXN][MAXN];
ll vis[MAXN][MAXN];

ll dx[4]={1,0,-1,0};
ll dy[4]={0,-1,0,1};

queue<arr> q;

inline void bfs(){	// 需要对之前所有加入q的点做好vis=true的处理
	while(!q.empty()){
		ll x=q.front()[0],y=q.front()[1],d=q.front()[2];
		q.pop();
		if(d>=D) continue;
		FOR(i,0,3){
			ll fx=x+dx[i],fy=y+dy[i];
			if(fx<1 || fx>N) continue;
			if(fy<1 || fy>M) continue;
			if(vis[fx][fy]<d+1) continue;
			if(maze[fx][fy]=='#') continue;
			vis[fx][fy]=d+1;
			q.push({fx,fy,d+1});
			// cerr<<fx<<' '<<fy<<'\n';
		}
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>N>>M>>D;
	FOR(i, 1, N){
		FOR(j,1,M){
			cin>>maze[i][j];
		}
	}
	CLR(vis,0x3f);
	ll stand=vis[0][0];
	FOR(i,1,N){
		FOR(j,1,M){
			if(maze[i][j]!='H') continue;
			q.push({i,j,0});
			vis[i][j]=0;
		}
	}
	bfs();
	ll ans=0;
	FOR(i,1,N){
		FOR(j,1,M){
			if(vis[i][j]!=stand) ++ans;
		}
	}
	cout<<ans<<endl;
	return 0;
}
/*
AC
https://atcoder.jp/contests/abc383/submissions/62366428
*/

心路历程（WA，TLE，MLE……）