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

A seaport container terminal stores large containers that are eventually loaded on seagoing ships for transport abroad. Containers coming to the terminal by road and rail are stacked at the terminal as they arrive.

Seagoing ships carry large numbers of containers. The time to load a ship depends in part on the locations of its containers. The loading time increases when the containers are not on the top of the stacks, but can be fetched only after removing other containers that are on top of them.

The container terminal needs a plan for stacking containers in order to decrease loading time. The plan must allow each ship to be loaded by accessing only topmost containers on the stacks, and minimizing the total number of stacks needed.

For this problem, we know the order in which ships must be loaded and the order in which containers arrive. Each ship is represented by a capital letter between A and Z (inclusive), and the ships will be loaded in alphabetical order. Each container is labeled with a capital letter representing the ship onto which it needs to be loaded. There is no limit on the number of containers that can be placed in a single stack.

Input 

The input file contains multiple test cases. Each test case consists of a single line containing from 1 to 1000 capital letters representing the order of arrival of a set of containers. For example, the line ABAC means consecutive containers arrive to be loaded onto ships A, B, A, and C, respectively. When all containers have arrived, the ships are loaded in strictly increasing order: first ship A, then ship B, and so on.

A line containing the word end follows the last test case.

Output 

For each input case, print the case number (beginning with 1) and the minimum number of stacks needed to store the containers before loading starts. Your output format should be similar to the one shown here.

Sample Input 

A 
CBACBACBACBACBA 
CCCCBBBBAAAA 
ACMICPC 
end

Sample Output 

Case 1: 1
Case 2: 3
Case 3: 1
Case 4: 4

---

題目描述：

希望有多個堆疊，去承載到對岸並且會由小到大。

題目解法：

即用最小的 LDS 去覆蓋全部的數列個數。

解法一是進行 DAG 圖上的最少路徑覆蓋，需要使用 matching 的演算法。

但事實上求 LIS 就是答案。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int main() {
    char s[1005];
    int i, j, cases = 0;
    while(gets(s)) {
        if(!strcmp(s, "end"))
            break;
        int dp[1005] = {}, n = strlen(s);
        int ret = 0;
        for(i = 0; i < n; i++) {
            for(j = i+1; j < n; j++) {
                if(s[j] > s[i])
                    dp[j] = max(dp[j], dp[i]+1);
            }
            ret = max(ret, dp[i]);
        }
        printf("Case %d: %d\n", ++cases, ret+1);
    }
    return 0;
}