Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem A: Trainsorting

Erin is an engineer. She drives trains. She also arranges the cars within each train. She prefers to put the cars in decreasing order of weight, with the heaviest car at the front of the train.

Unfortunately, sorting train cars is not easy. One cannot simply pick up a car and place it somewhere else. It is impractical to insert a car within an existing train. A car may only be added to the beginning and end of the train.

Cars arrive at the train station in a predetermined order. When each car arrives, Erin can add it to the beginning or end of her train, or refuse to add it at all. The resulting train should be as long as possible, but the cars within it must be ordered by weight.

Given the weights of the cars in the order in which they arrive, what is the longest train that Erin can make?

Input Specification

The first line is the number of test cases to follow. The test cases follow, one after another; the format of each test case is the following:

The first line contains an integer 0 <= n <= 2000, the number of cars. Each of the following n lines contains a non-negative integer giving the weight of a car. No two cars have the same weight.

Sample Input

1
3
1
2
3

Output Specification

Output a single integer giving the number of cars in the longest train that can be made with the given restrictions.

Output for Sample Input

3

---

做法 : DP

O(nlogn), max(LIS[i]+LDS[i]-1);

#include <stdio.h>
#include <stdlib.h>
void findLIS(int N, int LIS[], int pos[], int A[]) {
	int i, L = -1, l, r, m, newSet;
	int j;
	for(i = 0; i < N; i++) {
		l = 0, r = L, newSet = -1;
		while(l <= r) {
			m = (l+r)/2;
			if(pos[m] <= A[i]) {
				if(m == L || pos[m+1] > A[i]) {
					newSet = m+1;break;
				} else
					l = m+1;
			} else {
				r = m-1;
			}
		}
		if(newSet == -1)	newSet++;
		pos[newSet] = A[i];
		LIS[i] = newSet+1;
		if(L < newSet)	L = newSet;
	}
}
void findLDS(int N, int LDS[], int pos[], int A[]) {
	int i, L = -1, l, r, m, newSet;
	int j;
	for(i = 0; i < N; i++) {
		l = 0, r = L, newSet = -1;
		while(l <= r) {
			m = (l+r)/2;
			if(pos[m] >= A[i]) {
				if(m == L || pos[m+1] < A[i]) {
					newSet = m+1;break;
				} else
					l = m+1;
			} else {
				r = m-1;
			}
		}
		if(newSet == -1)	newSet++;
		pos[newSet] = A[i];
		LDS[i] = newSet+1;
		if(L < newSet)	L = newSet;
	}
}
int main() {
	int T, N, i;
	int A[2001], LDS[2001], LIS[2001], pos[2001];
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &N);
		for(i = 0; i < N; i++)
			scanf("%d", &A[N-i-1]);
		findLIS(N, LIS, pos, A);
		findLDS(N, LDS, pos, A);
		int max = 0;
		for(i = 0; i < N; i++) {
			if(LIS[i]+LDS[i]-1 > max) 
				max = LIS[i]+LDS[i]-1;
		}
		printf("%d\n", max);
	}
    return 0;
}