rainboy的解析

§ 解析

具体看代码

dp转移方程

dp[i][j]=max{dp[i1][j],dp[i1][j1]}+a[i][j] dp[i][j] = max\{dp[i-1][j] ,dp[i-1][j-1] \} + a[i][j]

dp[i][j]dp[i][j]表示从起点,到达第i行,第j列这个点后能得到的最大值

cpp
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#include <iostream> using namespace std; const int maxn = 1000 +5; int n; int a[maxn][maxn]; int f[maxn][maxn]; int ans; void init() { cin >> n; for(int i =1;i<=n ;i++){ for(int j =1;j<=i;j++) cin >> a[i][j]; } for(int i =0;i<=1000;i++) for (int j = 0; j <= 1000; j++) f[i][j] = -1; } // dfs(i,j) 表示从i,j这个点开始向下走,到 // 最后一层的最值 int dfs(int i,int j) { if( i == n+1) { return 0; } if( f[i][j] != -1) return f[i][j]; int x = dfs(i+1,j); int y = dfs(i+1,j+1); if( x > y) f[i][j] = x; else f[i][j] = y; f[i][j] += a[i][j]; return f[i][j]; } int main(int argc, char const *argv[]) { init(); dfs(1,1); cout << f[1][1] << endl; return 0; }