2 solutions

  • 1
    @ 2022-1-6 18:40:50
    #include<iostream>
    using namespace std;
    int n,m,s,ans=0;
    int ma[110][110];
    int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
    //传参的sum就是到(x,y)时经过的路径数字总和
    void dfs(int x,int y,int sum){
    	ans=max(ans,sum);
    	for(int i=0;i<4;i++){
    		int xx=x+dx[i];
    		int yy=y+dy[i];
    		if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&ma[xx][yy]>ma[x][y])
    			dfs(xx,yy,sum+ma[xx][yy]);
    	}
    }
    int main() {
    	cin>>n>>m>>s;
        //通过题目所给的算法建好地图
    	for(int i=1;i<=n;i++)
    		for(int j=1;j<=m;j++){
    			s=(s*345)%19997;
    			ma[i][j]=s%10+1;
    		}
        //把每个点都作为起点开始搜索
    	for(int i=1;i<=n;i++)
    		for(int j=1;j<=m;j++)
    			dfs(i,j,ma[i][j]);
    	cout<<ans<<endl;
    	return 0;
    }
    
    • 0
      @ 2025-1-31 14:16:22

      #include<bits/stdc++.h> using namespace std; const int N=200; int n,m,s,a[N][N],cnt=0; int dx[4]={0,-1,0,1}; int dy[4]={-1,0,1,0}; struct point{ int x; int y; int nums; }; queueq; void bfs(int x,int y) { point temp; temp.x=x; temp.y=y; temp.nums=a[x][y]; q.push(temp); while(!q.empty()) { temp=q.front(); cnt=max(cnt,temp.nums); for(int i=0;i<4;i++){ int nx=temp.x+dx[i]; int ny=temp.y+dy[i]; if(a[nx][ny]>a[temp.x][temp.y]&&nx>=0&&nx<n&&ny>=0&&ny<m){ q.push({nx,ny,temp.nums+a[nx][ny]}); } } q.pop(); } } int main() { int i,j; cin>>n>>m>>s; for(i=0;i<n;i++){ for(j=0;j<m;j++){ s=(s*345)%19997; a[i][j]=s%10+1; } } for(i=0;i<n;i++){ for(j=0;j<m;j++){ bfs(i,j); } } cout<<cnt; return 0; }

      • 1

      Information

      ID
      718
      Time
      1000ms
      Memory
      128MiB
      Difficulty
      4
      Tags
      # Submissions
      110
      Accepted
      48
      Uploaded By