Morris' Blog 我在摸索我存在的意、生命的意、涵及值。 有真正神手般的害，套用模式是大部分的情。 那害的我存在的意何，托出神的存在？ 有候不能怪人不我，我已始省思， 什我有人的原因，而是全要靠人。 有候我我不同，那是因我做不到很多人能到的事情。 我不到、玩不起、得不明、理解得不多， 我缺少的文能力，就是一切一切能的判定。 此，我不得能做些人有助的事情。 我知道我自己是不服的性格，但在我不得不。 我有能力，因此我必。 我到底世界作什？充不定性， 而我是法出拔萃，想一般人似乎也回不了了。 越大的挑需要越力的支持，而我的支持是什？ 最近人，但都是失的局，人能接受， 不需要言，不需要通，一切所想都表不出。

49的鼓 6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;
}