Codeforces round #262 (Div. 2) C (Binary answer, latency tag)

This is a type of question that maximizes the minimum value. This type of question is typically answered using a bipartite enumeration.

In the case of a binary answer, first determine the exact range of the answer. Then, the key is how to judge the answer line for each enumeration.

I use the [I] array to indicate the height of the flowers in the initial phase. B [I] indicates how much the flowers will increase to reach the answer of the current enumeration (that is, the value of the MID. The two arrays are very computation. The key is the continuous pouring of W flowers. How to update the interval (violent O (n2) Will time out after the update )? You can use a line segment tree, but this question does not involve a range query.Update intervals on an array, Using the line segment tree. In fact, this update can be solved by O (n) with the concept of delay mark (lazy operation. For more information, see hdu4970.

Lazy [] the function of marking the array is to record where the update interval is stopped.

# Include <iostream> # include <cstdio> # include <cstdlib> # include <cstring> # include <cmath> # include <map> # include <set> # include <vector> # include <algorithm> # include <stack> # include <queue> using namespace STD; # define INF 1000000000 # define EPS 1e-8 # define PII pair <int, int> # define ll long intconst int maxn = 110009; int n, m, W, a [maxn], B [maxn], lazy [maxn], ANS = 0;/* lazy [] the function of marking the array is actually to record where to stop the update range */INT main () {// freopen ("in6.txt", "r", stdin); scanf ("% d", & N, & M, & W ); for (INT I = 1; I <= N; I ++) {scanf ("% d", & A [I]) ;} int L = 1, R = maxn + INF, mid; while (L <= r) {memset (lazy, 0, sizeof (lazy); Mid = (R + l)/2; for (INT I = 1; I <= N; I ++) B [I] = max (0, mid-A [I]); int TM = m, X = 0; For (INT I = 1; I <= N; I ++) // start watering {x + = lazy [I]; // check whether this step is marked with B [I]-= x; // Add the watering operation that affects the current B [I] To if (B [I]> 0) // It must be greater than 0, then you have to start watering it again {TM-= B [I]; If (tm <0) break; // The number of days is limited to lazy [I + W]-= B [I]; x + = B [I] ;}} if (tm <0) r = mid-1; else {ans = mid; L = Mid + 1 ;}} printf ("% d \ n", ANS );}

 

Codeforces round #262 (Div. 2) C (Binary answer, latency tag)