NC23486 小A与小B(双向BFS)
小A与小B
https://ac.nowcoder.com/acm/problem/23486
题意:
题解:
AC代码
/*
Author:zzugzx
Lang:C++
Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[8][2]={{0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
int n,m;
bool vis[2][1010][1010];
queue<pii> q[2];
char g[1010][1010];
bool bfs(int x){
int sz=SZ(q[x]);
while(sz--){
pii p=q[x].front();
q[x].pop();
for(int k=0;k<(x?4:8);k++){
int dx=p.fi+dir[k][0];
int dy=p.se+dir[k][1];
if(dx>=1&&dx<=n&&dy>=1&&dy<=m&&g[dx][dy]!='#'&&!vis[x][dx][dy]){
if(vis[x^1][dx][dy])return 1;
vis[x][dx][dy]=1;
q[x].push(mp(dx,dy));
}
}
}
return 0;
}
int solve(){
int res=0;
while(!q[0].empty()||!q[1].empty()){
res++;
if(bfs(0))return res;
if(bfs(1))return res;
if(bfs(1))return res;
}
return -1;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++){
cin>>g[i][j];
if(g[i][j]=='C')q[0].push(mp(i,j));
if(g[i][j]=='D')q[1].push(mp(i,j));
}
int ans=solve();
if(ans==-1)cout<<"NO";
else {
cout<<"YES"<<endl;
cout<<ans<<endl;
}
return 0;
}
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