6 solutions

  • 1
    @ 2023-8-17 9:53:38

    动态规划,来看我的。

    #include<bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    //dp[i][j]//i:高度,j:位置
    /*
    dp[i][j] = max(dp[i+1][j],dp[i+1][j]) + w[i][j]
    
     */
    const int N = 1e3+10;
    int w[N][N]; //每个点的权值
    LL dp[N][N]; //存储每个点及以下的权值和
    int main(){
    	int R;
    	cin>>R;
    	for(int i=1;i<=R;i++){
    		int j=i;
    		for(int j=1;j<=i;j++){
    			int c;
    			cin>>c;
    			w[i][j] = c;
    		}
    	}
    	for(int i=R;i>=1;i--){
    		for(int j=1;j<=i;j++){
    			if(i==R){
    				dp[i][j]=w[i][j];
    			}
    			else{
    				dp[i][j] = max(dp[i+1][j],dp[i+1][j+1]) + w[i][j];
    			}
    		}
    	}
    	cout<<dp[1][1]<<endl;
    	return 0;
    }
    

    Information

    ID
    87
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    5
    Tags
    # Submissions
    223
    Accepted
    91
    Uploaded By