Given Round E to go to the ... miss this village no this shop so tidy 0.0
Here plus an outline. markdown how to use the blog garden ......
Well format in this first issue come. . and have goals qwq. !
## Topic 2 Circuit Board
The 84 on LeetCode is a classic problem. LeetCode I still just do .. anyway very little, nor took Hard, easy feeling is always the water of the past, medium is also very little.
The next step might be maintained on github what? A separate sub-problem solutions and separate explanations ...... can write then transplanted in the past on screen. (Mixed feelings .jpg) may not transplant
### Italian title
300 * 300 matrix, given a lot number, so that the maximum value in each row, does not exceed the minimum value K, a minimal matrix satisfies these conditions is.
### false (false most of the deleted ......)
o (n ^ 3) approach.
Began to think that the disjoint-set ownership, strings of children as chatter in the past, but actually a bit emmmm
For such cases, only to find the individual continuous then began to spread from the individual is not enough.
2 2 2 4
4 2 2 2
2 2 2 4
Feeling that you need to put aside racial prejudice, can not be so narrow-minded to think about, and n is small, you can directly enumerate two-dimensional, one-dimensional again
First, let's use a three-dimensional Boolean array to record, f [i] [l] [r], is recorded at the i-th row from l to r, the condition can achieve in this segment, "no more than E" is.
Here is a very rigid, even without considering the maximum value if there are two problems, just pretreatment, enumeration l and r.
it is good. If the Boolean type has been determined
### boolean how to update?
I thought, or to update at least n * (n-1) / 2 th.
That is really funny. If the ratio is now front and no cross-border size,
I am currently in position a2, the current has been matched to the len (forget a2), then 1 ~ a2-len fill No, a2-len + 1 ~ a2 fill Yes (true).
How to see the size of it out of bounds?
Ah, this also seemed to have time-sensitive, the most stupid way, it is better to meet the time being to come and K, are not satisfied when 1 ~ a2-len + 1 still can, while walking back and tmpmin maintenance tmpmax
Think.
If l and r directly maintain it?
Ah, here, the gap between thinking manifested. Our I wrote in the above methods may be possible, but cumbersome, and there is no sense that there is a "specialized" feeling.
Despite these Think again, at the end of this article, there is provided a one-dimensional answer how can nlogn, but not necessary.
This process is separate from the main program and write, can write separate enumeration, which means that ...
Direct violence l and r, even if n ^ 2 directly run again, each time look called "l to r" range, just write a false or true. Even so complexity is the square of n is sufficient
Stripping out a unified look, thinking to rise a level. .
### true =================================
Oh, I feel so stupid, you can verify the above are false and do not have the height of the tube.
Its written explanations, or are you stroke along the best of it ......
……………………
This problem is only a word. Enumeration l and r, then the width of the string is determined, l and r have the n ^ 2
After determining the width, in the end how long, and directly determines yes no, then use the cumulative and multiplied by r-l + 1 of the can.
Why should I do? To verify the problem, said the longest block, then the length (sideways) OK, go on end line and look better, or is bristling OK, it sideways line and a string of best
Obviously, the length should be OK, and then see a few lines can meet this condition. Direct line by line to see it (Figure 1)
If the width is determined, even if is to continue to expand, it is useless, as not to make this distinction, we can see 113 Thus, according to this division does not mean that, because this is not the intention of the subject.
(Figure II)
Then write the code. Tired ah fight resolve to fight hand hurt I do not want to write (buxing
Write an example:
99340
99340
2 2 2 3
3 2 2 2
2 2 2 3
Hand ...... after 26 minutes to sort out ideas, is no problem, but I was wrong a ==, one is not updated when the 1,1 This update into the header
The key is thinking ah! ! ! This is a very simple ah! ! ! ! 26 points ah! ! ! ! kickstart miss gone ah ah ah ah! ! !
Finally, stick the code
#include <the iostream> #include <algorithm> #include <Queue> the using namespace STD; int n-; BOOL F [ 305 ] [ 305 ] [ 305 ]; int A [ 305 ] [ 305 ]; int A1, A2, K; int main () { int T; CIN >> T; int kase = 0 ; the while (T-- ) { // F if not enough empty initialized directly // input CIN >> >> A1 A2 >> K; for (int i = 1; i <= a1; i++) { for (int j = 1; j <= a2; j++) { cin >> a[i][j]; } } //对数据处理 初始化bool f bool ff; int now; for (int i = 1; i <= a1; i++) {//行 for (int left= 1; left <= a2; left++) { //下面的列 f[i][left][left] = true; int maxx = a[i][left]; int= Minn A [I] [left]; // [shock! I am here to write the equal sign] // [scared! 2 I forgot to take into account the first one! Yeah] FF = to true ; for ( int right = left + . 1 ; right <= A2; right ++ ) { IF (FF == to false ) F [I] [left] [right] = to false ; the else { now = A [I ] [right]; Maxx = max (Maxx, now); Minn = min (Minn, now); IF (Minn + K <K-Maxx now ||> now) FF = to false ; f [i] [left] [right] = FF; } // this number and compare minn maxx like at the same time update the look // If you have a false overall it is a false good L } } } / * for (int i = 1; i < = a1; i ++) {// row for (int left = 1; left <= a2; left ++) {// following column for (int right = left; right <= a2; right ++ ) { COUT << F [I] [left] [right] << ""; } COUT << endl; } COUT << endl; } * / int maxans = - . 1 ; int ANS = 0 ; // .. well .. that deal with the end of the cycle and the last one left to shoot for (int l = 1; l <= a2; l++) { for (int r = l; r <= a2; r++) { ans = 0; for (int i = 1; i <= a1; i++) { if (f[i][l][r] == true) { ans += (r - l + 1); maxans = max(ans, maxans); } else { ans = 0; } } } } cout <<"Case #"<<++kase<<": "<< maxans << endl; } return 0; }
### Deformation
But to say this topic is a famous deformation, as well:
this is a schematic diagram of hungry ... .. meaning the equivalent of the subject so let us seek. In fact, it can be transformed into such topics leetcode of:
this is a schematic diagram of hungry ... .. meaning the equivalent of the subject so let us seek. In fact, it can be transformed into such topics leetcode of:
### Extended problem:
If #### is one-dimensional?
It looks directly sideways past the line. They can then add, not the throw
There is a problem, the maximum value is the minimum value, if it is 345,727 came back and 1, it is not finished yet, a simple double pointer ineffective.
Cheating a little practice: with mutiset.
But also to s.erase (s.find (7)) would only remove one, two or gone 7
Look largest long s.rbegin () can be. On cross-border move forward (deletion), not on cross-border security can be achieved nlogn complexity.
If #### is three-dimensional?
Imagine, in fact, n ^ 5 to n ^ 4 to limit the length, the equivalent of a rectangular bucket, continue to meet, it seems a bit cumbersome pretreatment