2 solutions

  • 0
    @ 2022-3-20 19:42:02

    bfs不太会,我强行用了dfs

    #include<bits/stdc++.h>
    using namespace std;
    int n,m,x,y,sx,sy,fx,fy;
    char ma1[25][25];
    int ans[10005];//记录每个到达终点的cnt的值
    bool ma2[25][25];//记录是否走过 
    int opx[5] = {1,0,0,-1};
    int opy[5] = {0,1,-1,0};
    int cnt,op=0;//cnt 用于计数 ans 用于记录最小的路径 
    void dfs(int x,int y){
    	if(x == fx && y == fy){
    		ans[op] = cnt;
            op++;
            
    	}
    	else{
    		for(int i = 0; i < 4; i++){
    			int nex = x + opx[i];
    			int ney = y + opy[i];
    			if(nex>=1&&nex<=n&&ney>=1&&ney<=m&&(ma1[nex][ney]=='.'||ma1[nex][ney]=='*')&&!ma2[nex][ney]){
    				ma2[nex][ney] = true;
    				cnt++; 
    				dfs( nex, ney);
    				ma2[nex][ney] = false;
                    cnt--;
    			}
    		}
    	}
    }
    int main(){
    	std::ios::sync_with_stdio(false);
    	cin>>n>>m;
    	for(int i = 1; i <= n; i++){
    		for(int j = 1; j <= m; j++){
    			cin>> ma1[i][j];
    			if(ma1[i][j]=='@')
    				sx = i,sy = j;
    			if(ma1[i][j]=='*')
    				fx = i, fy = j;
    		}
    	}
    	dfs( sx, sy);
        int mini=100005;
        for(int i=0;i<100;i++){
            if(ans[i]&&ans[i]<mini)
                mini=ans[i];
        }
        if(mini != 100005)
    	    cout<<mini;
       else
           cout<<-1;
    	return 0;
    }
    
    • 0
      @ 2022-3-20 14:48:40

      吐槽一下关于周日不算本周的事(不要捶我),好的本题是一道bfs的板子题(大概),上代码(献丑了)

      #include<iostream>
      #include<queue>
      using namespace std;
      const int N=27;
      bool v[N][N];
      int n,m,flag=0;
      char a[N][N];
      int dx[4]={-1,0,1,0},dy[4]={0,1,0,-1};
      struct A{
      	int step,x,y;
      }p[N*N];
      queue<A>q;
      int main(){
      	cin>>n>>m;
      	for(int i=1;i<=n;i++){
      		for(int j=1;j<=m;j++){
      			cin>>a[i][j];
      			if(a[i][j]=='@'){
      				v[i][j]=true;
      				A sta;
      				sta.step=0;
      				sta.x=i;
      				sta.y=j;
      				q.push(sta);
      			}
      		}
      	}
      	while(!q.empty()){
      		A t=q.front();
      		v[t.x][t.y]=true;
      		if(a[t.x][t.y]=='*'){
      			cout<<t.step;
      			flag=1;
      			break;
      		}
      		q.pop();
      		for(int i=0;i<4;i++){
      			int xx=t.x+dx[i];
      			int yy=t.y+dy[i];
      			if(!v[xx][yy]&&xx>0&&yy>0&&xx<=n&&yy<=m&&a[xx][yy]!='#'){
      				A next;
      				next.x=xx;
      				next.y=yy;
      				next.step=t.step+1;
      				q.push(next);
      			}
      		}
      	}
      	if(flag==0)cout<<"-1";
      	return 0;
      }
      
      • 1

      Information

      ID
      1236
      Time
      1000ms
      Memory
      128MiB
      Difficulty
      5
      Tags
      # Submissions
      99
      Accepted
      38
      Uploaded By