2022 the first of the new league henan (4) : zhengzhou university of light industry G - maze

G - 迷宫

最短路问题,Built two edge,If the target point for spend so edge right for $1-1=0$ ,否则为 $1$ Fixed source point and the sink,The short circuit of various algorithms can write,This problem is written in deque is better,But I was using a priority queue.

优先队列

#include<bits/stdc++.h>
using namespace std;

typedef pair<int,int> PII;
typedef pair<int,PII> PIII;

const int N = 2010;
 
char g[N][N];
int f[N][N];
int n,m;

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

void bfs()
{
    
	memset(f,0x3f,sizeof f);
    f[1][1]=0;
    priority_queue<PIII,vector<PIII>,greater<PIII> > Q;
    Q.push({
    0,{
    1,1}});
     
    while(Q.size())
    {
    
        PIII t=Q.top();Q.pop();
        for(int i=0;i<4;i++)
        {
    
            int x=t.second.first+dx[i],y=t.second.second+dy[i];
            if(x<1||x>n||y<1||y>m) continue;
            if(g[x][y]=='#') continue;
            if(f[x][y]!=0x3f3f3f3f) continue;
            if(g[x][y]=='.') f[x][y]=min(f[x][y],f[t.second.first][t.second.second]+1);
            if(g[x][y]=='*') f[x][y]=min(f[x][y],f[t.second.first][t.second.second]);
            Q.push({
    f[x][y],{
    x,y}});
        }
    }
}

signed main()
{
    
	cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>(g[i]+1);
	
	bfs();
	
	if(f[n][m]>=0x3f3f3f3f/2) cout<<"-1"<<endl;
	else cout<<f[n][m]<<endl;
    return 0;
}

双端队列

#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int> PII;

const int N = 2010;
 
char g[N][N];
int f[N][N];
int n,m;

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

void bfs()
{
    
	memset(f,0x3f,sizeof f);
	f[1][1]=0;
	deque<PII> Q;
	Q.push_front({
    1,1});
	
	while(!Q.empty())
	{
    
		PII t=Q.front();Q.pop_front();
		for(int i=0;i<4;i++)
		{
    
			int tx=t.first+dx[i],ty=t.second+dy[i];
			if(tx<1||tx>n||ty<1||ty>m) continue;
			if(g[tx][ty]=='#') continue;
			int d=(g[tx][ty]=='.');
			if(f[tx][ty]>f[t.first][t.second]+d)
			{
    
				f[tx][ty]=f[t.first][t.second]+d;
				if(d==0) Q.push_front({
    tx,ty});
				else Q.push_back({
    tx,ty});
			}
		}
	}
}

int main()
{
    
	cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>(g[i]+1);
	
	bfs();
	
	if(f[n][m]>=0x3f3f3f3f/2) cout<<"-1"<<endl;
	else cout<<f[n][m]<<endl;
	return 0;
}