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

During this economic crisis time, Jack has started an incredible new business related to air travel, a parking-lot for airplane. He bought a very large land to park airplanes. However the land is very narrow, so that the only way airplanes can go in or go out of the parking lot must be in the Last-In First-Out fashion (see picture below). He only has spaces in the parking lot so he cannot take some airplane at the end out so that other airplanes can move.

Because of the limitation of the parking lot, it is not possible to accommodate all requests for parking. Each request consists of the planned arrival time and planned departure time, which are the times the airplane arrives at the parking lot. An example below shows a request table for 4 planes.

In this case, it is possible to accommodate airplane 1, 2, and 4. But it is not possible to accommodate both airplanes 2 and 3.

It is possible that different planes plan to arrive or depart the parking lot at the same time. Jack has the best crews working with him, so that they will manage to arrange the plane to the parking lot in the best way that if it is possible to park and take out the planes they will be able to do it. Consider another example.

Although airplane 5 and 6 arrive at the same time, Jack's crews know that airplane 5 will have to be out before airplane 6, so when both airplanes arrive they put airplane 6 in first, following by airplane 5.

Given a list of parking requests, you want to find the maximum number of airplanes that can be parked in this parking lot, provided that they can only depart in the Last-In First-Out fashion.

Input 

The first line contains an integer T, the number of test cases (1T5). Each test case is in the following format.

The first line starts with an integer N (1N300) denoting the number of airplanes. The next N lines describe the request table. Each line 1 + i, for 1iN, contains two integer S_i and T_i , (0S_i < T_i1, 000, 000, 000) which are the planned arrival time and planned departing time for airplane i.

Output 

For each test case, you program must output a single line consisting of one integer, the maximum number of airplanes that can be parked in Jack's parking lot.

Sample Input 

2 
4 
1 10 
2 5 
3 7 
6 9 
3 
10 12 
10 15 
13 17

Sample Output 

3 
2

---

把時間換成括弧，但括弧可以重疊。
有點矩陣鏈乘積的做法，消耗 O(N^3)，尚未學到四邊形不等式，暫無優化。

#include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int t, n, tn;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int S[305], T[305], i, j, k;
        map<int, int> r;
        for(i = 0; i < n; i++) {
            scanf("%d %d", &S[i], &T[i]);
            r[S[i]] = 1;
            r[T[i]] = 1;
        }
        i = 0;
        for(map<int, int>::iterator it = r.begin();
            it != r.end(); it++) {
            it->second = i;
            i++;
        }
        tn = i+1;
        int dp[601][601] = {};
        char has[601][601] = {};
        for(i = 0; i < n; i++) {
            has[r[S[i]]][r[T[i]]] = 1;
        }
        int ans = 1;
        for(i = 0; i < tn; i++) {
            for(j = 0; j+i < tn; j++) {
                dp[j][j+i] = 0;
                if(i)
                    dp[j][j+i] = max(dp[j+1][j+i], dp[j][j+i-1]);
                for(k = j; k <= j+i; k++) {
                    dp[j][j+i] = max(dp[j][j+i], dp[j][k]+dp[k][j+i]);
                }

                dp[j][j+i] += has[j][j+i];
                if(dp[j][j+i] > ans)
                    ans = dp[j][j+i];
            }
        }
        cout << ans << endl;
    }
    return 0;
}