AcWing 175. 电路维修 BFS+双端队列

Posted 2021-11-11 karshey

题

注意边界：

/*
参考题解：
https://www.acwing.com/solution/content/26973/ 
https://www.acwing.com/solution/content/21775/
观看视频：
https://www.acwing.com/video/132/ 
*/ 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb push_back
#define fi first
#define se second
#define mem(a,x) memset(a,x,sizeof(a));
#define db double 
#define fir(i,a,n) for(int i=a;i<=n;i++)
#define debug(x) cout<<#x<<" "<<x<<endl;
const int inf=0x3f3f3f3f;
const int MOD=1e9+7;
//要注意边界范围======================
const int N=510;
int n,m;
char g[N][N];
int ans[N][N];
deque<pii>dq;
int dx[4]={-1,-1,1,1};
int dy[4]={-1,1,-1,1};//格点的四个方向 
int ix[4]={-1,-1,0,0};
int iy[4]={-1,0,-1,0}; 
string idxy="\\\\//\\\\";
void bfs()
{
	dq.clear();
	mem(ans,0x3f);
	dq.push_back({0,0});
	ans[0][0]=0;
	while(dq.size())
	{
		auto t=dq.front();
		dq.pop_front();
		
		for(int i=0;i<4;i++)
		{
			int x=t.fi+dx[i];
			int y=t.se+dy[i];
			
			if(x>=0&&x<=n&&y>=0&&y<=m)
			{
				int w=0;//先假设权值为0
				int xx=t.fi+ix[i];
				int yy=t.se+iy[i];
				if(g[xx][yy]!=idxy[i]) w=1;
				if(ans[x][y]>ans[t.fi][t.se]+w)
				{
					ans[x][y]=ans[t.fi][t.se]+w;
					if(w) dq.push_back({x,y});
					else dq.push_front({x,y});
				}
			}
		}
	}
}
int main()
{
	int t;cin>>t;
	while(t--)
	{
		cin>>n>>m;
		for(int i=0;i<n;i++) cin>>g[i];
		bfs();
		if(ans[n][m]==0x3f3f3f3f) cout<<"NO SOLUTION";
		else cout<<ans[n][m];
		cout<<endl;
	}
	return 0; 
}