2 solutions
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1
#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; }
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0
#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