RockyB's blog

By RockyB, 2 years ago, translation, , Hi CodeForces!

In this blog written that in problems Divide and Conquer Optimization opt[i] <= opt[i + 1]

opt[i] -> such k that gives optimal answer, for example dp[i] = dp[k - 1] + C[k][i]

Please, can you explain why? Comments (2)
 » 2 years ago, # | ← Rev. 3 →   That monotonicity condition is what allows us to use Divide and Conquer optimization to reduce O(N^2) to O(N*logN). Take a look at some sample code. We call solve(L, R, optima_l, optima_r) to find the optimal split points for all indices i such that L <= i <= R, subject to the condition that optima_l <= opt[i] <= optima_r. We achieve this by finding the optimal point for mid=(L+R)/2 by iterating in the range [optima_l, optima_r]. Then we call solve(L, mid-1, optima_l, optima_mid) and solve(mid+1, R, optima_mid, optima_r) recursively. There are logN levels of recursion. Consider a single level. The TOTAL number of iterations over all intervals in that level is O(n). This is because opt <= opt <= opt .... <= opt[N]. So this monotonicity condition is the reason we can reduce the time complexity in this manner. Sometimes, formally proving the monotonicity condition can become difficult for a problem, maybe during an ongoing contest. In that case, I mostly rely on intuition or use another condition (which is sufficient but not necessary) to apply this optimization. I have found this second quadrangular inequality condition to be applicable to many problems. :)
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