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 packet-switching network handles information in small units, breaking long messages into multiple packets before routing. Although each packet may travel along a different path, and the packets comprising a message may arrive at different times or out of order, the receiving computer reassembles the original message correctly.

The receiving computer uses a buffer memory to hold packets that arrive out of order. You must write a program that calculates the minimum buffer size in bytes needed to reassemble the incoming messages when the number of messages (N ), the number of packets (M ), the part of the messages in each packet, the size of each message, and the order of the incoming packets are given.

When each packet arrives, it may be placed into the buffer or moved directly to the output area. All packets that are held in the buffer are available to be moved to the output area at any time. A packet is said to ``pass the buffer" when it is moved to the output area. A message is said to ``pass the buffer" when all of its packets have passed the buffer.

The packets of any message must be ordered so the data in the sequence of packets that pass the buffer is in order. For example, the packet containing bytes 3 through 5 of a message must pass the buffer before the packet containing bytes 6 through 10 of the same message. Messages can pass the buffer in any order, but all packets from a single message must pass the buffer consecutively and in order (but not necessarily at the same time). Note that unlike actual buffering systems, the process for this problem can look ahead at all incoming packets to make its decisions.

The packets consist of data and header. The header contains three numbers: the message number, the starting byte number of data in the packet, and the ending byte number of data in the packet respectively. The first byte number in any message is 1.

For example, the figure below shows three messages (with sizes of 10, 20, and 5 bytes) and five packets. The minimum buffer size for this example is 10 bytes. As they arrive, packet #1 and packet #2 are stored in the buffer. They occupy 10 bytes. Then packet #3 passes the buffer directly. Packet #4 passes the buffer directly and then packet #2 exits the buffer. Finally, packet #5 passes the buffer directly and packet #1 exits the buffer.

Input 

The input file contains several test cases. The first line of each test case contains two integers N (1N5) and M (1M1000) . The second line contains N integers that are the sizes of messages in bytes; the first number denotes the size of message #1, the second number denotes the size of message #2, and so on. Each of the following M lines describes a packet with three integers: the message number and the starting and ending byte numbers of data in the packet. No packet contains more 64 bytes of data.

The last test case is followed by a line containing two zeroes.

Output 

For each test case, print a line containing the test case number (beginning with 1) followed by the minimum buffer size in bytes required to reassemble the original messages. Put a blank line after the output for each test case. Use the format of the sample output.

Sample Input 

3 3 
5 5 5 
1 1 5 
2 1 5 
3 1 5 
3 5 
10 20 5 
2 16 20 
1 6 10 
3 1 5 
1 1 5 
2 1 15 
0 0

Sample Output 

Case 1: 0 

Case 2: 10

---

題目解法：
題目給定每個封包不超過 64 bytes => 得到 message 不超過 64000 bytes
由於要自己決定哪個封包要先處理，因此要窮舉 n!，然後在模擬整個過程。

題目說明訊息要依序完成，而每個訊息內部的封包要依序完成。
那麼基於 greedy 的想法，能從 buffer 清出就盡量清出，最後比較其最大的 buffer size 中的最小。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
int buffer[5][65536][2] = {};
int main() {
    int n, m, cases = 0;
    int len[10], i;
    while(scanf("%d %d", &n, &m) == 2 && n) {
        int A[10];
        for(i = 0; i < n; i++) {
            scanf("%d", &len[i]);
            A[i] = i;
        }
        int msg[1005], l[1005], r[1005];
        int ret = 0xfffffff;
        for(i = 0; i < m; i++) {
            scanf("%d %d %d", &msg[i], &l[i], &r[i]);
            msg[i]--;
        }
        do {
            for(i = 0; i < n; i++)
                memset(buffer[i], 0, (len[i]+10)*4*2);
            int bufsize = 0, sub = 0;
            int order[5] = {1,1,1,1,1};
            int accept = 0;
            for(i = 0; i < m; i++) {
                int ll = l[i], rr = r[i], kmsg = msg[i];
                if(order[kmsg] == ll && kmsg == A[accept]) {
                    order[kmsg] = rr+1;
                    while(buffer[kmsg][order[kmsg]][0]) {
                        bufsize -= buffer[kmsg][order[kmsg]][1]-order[kmsg];
                        order[kmsg] = buffer[kmsg][order[kmsg]][1];
                    }// pop this message
                    while(accept+1 < n && order[A[accept]] == len[A[accept]]+1) {//pop next message
                        accept++;
                        int kmsg = A[accept];
                        while(buffer[kmsg][order[kmsg]][0]) {
                            bufsize -= buffer[kmsg][order[kmsg]][1]-order[kmsg];
                            order[kmsg] = buffer[kmsg][order[kmsg]][1];
                        }
                    }
                } else {
                    bufsize += rr-ll+1;
                    buffer[kmsg][ll][0] = 1;
                    buffer[kmsg][ll][1] = rr+1;
                }
                sub = max(sub, bufsize);
                if(sub >= ret)  break;//impossible
            }
            ret = min(ret, sub);
        } while(next_permutation(A, A+n));
        printf("Case %d: %d\n\n", ++cases, ret);
    }
    return 0;
}