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 B: Ferry Loading III

Before bridges were common, ferries were used to transport cars across rivers. River ferries, unlike their larger cousins, run on a guide line and are powered by the river's current. Cars drive onto the ferry from one end, the ferry crosses the river, and the cars exit from the other end of the ferry.

There is a ferry across the river that can take n cars across the river in t minutes and return in t minutes. A car may arrive at either river bank to be transported by the ferry to the opposite bank. The ferry travels continuously back and forth between the banks so long it is carrying a car or there is at least one car waiting at either bank. Whenever the ferry arrives at one of the banks, it unloads its cargo and loads up to n cars that are waiting to cross. If there are more than n, those that have been waiting the longest are loaded. If there are no cars waiting on either bank, the ferry waits until one arrives, loads it (if it arrives on the same bank of the ferry), and crosses the river. At what time does each car reach the other side of the river?

The first line of input contains c, the number of test cases. Each test case begins with n, t, m. m lines follow, each giving the arrival time for a car (in minutes since the beginning of the day), and the bank at which the car arrives ("left" or "right"). For each test case, output one line per car, in the same order as the input, giving the time at which that car is unloaded at the opposite bank. Output an empty line between cases.

You may assume that 0 < n, t, m ≤ 10000. The arrival times for each test case are strictly increasing. The ferry is initially on the left bank. Loading and unloading time may be considered to be 0.

Sample input

2
2 10 10
0 left
10 left
20 left
30 left
40 left
50 left
60 left
70 left
80 left
90 left
2 10 3
10 right
25 left
40 left

Output for sample input

10
30
30
50
50
70
70
90
90
110

30
40
60

---

Gordon V. Cormack

---

描述一下題目內容

渡船一開始在左側，
根據時間，車子依序抵達。
如果兩邊都沒有車子，渡船就會在那一側持續等待。
然而，如果一邊有車子，而渡船那一側還沒有車子，渡船則會開到另一邊去載。

如果渡船抵達後，有可以載的車子，最多載 N 輛，便立刻出發到另一岸。

結果 ... 代碼被我寫醜了，因為我看不太懂題目。

#include <stdio.h>
#include <queue>
#include <algorithm>
using namespace std;
struct car {
    int t, in;
    car(int x, int y) {
        t = x;
        in = y;
    }
};
int main() {
    int testcase, n, m, t, i;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d %d %d", &n, &t, &m);
        int time;
        char s[50];
        queue<car> LEFT, RIGHT;
        for(i = 0; i < m; i++) {
            scanf("%d %s", &time, s);
            if(s[0] == 'l')
                LEFT.push(car(time, i));
            else
                RIGHT.push(car(time, i));
        }
        int now_time = 0, flag = 0;
        int res[10000];
        while(1) {
            if(LEFT.empty() && RIGHT.empty())
                break;
            //printf("time %d : %s\n", now_time, flag ? "right" : "left");
            int tn = 0;
            if(flag == 0) { // LEFT bank
                while(tn < n && !LEFT.empty() && LEFT.front().t <= now_time)
                    tn++, res[LEFT.front().in] = now_time+t, LEFT.pop();
                if(tn) { // goto either
                    now_time += t;
                    flag = 1-flag;
                    continue;
                }
                if(RIGHT.empty() || (!LEFT.empty() && LEFT.front().t <= RIGHT.front().t)) {
                    now_time = LEFT.front().t;// waiting left
                    while(tn < n && !LEFT.empty() && LEFT.front().t <= now_time)
                        tn++, res[LEFT.front().in] = now_time+t, LEFT.pop();
                    now_time += t;
                } else {
                    if(!RIGHT.empty() || LEFT.empty() && LEFT.front().t > RIGHT.front().t)
                        now_time = max(RIGHT.front().t, now_time)+t;//max tricky
                    else
                        now_time += t;
                }
            } else {
                while(tn < n && !RIGHT.empty() && RIGHT.front().t <= now_time)
                    tn++, res[RIGHT.front().in] = now_time+t, RIGHT.pop();
                if(tn) {
                    now_time += t;
                    flag = 1-flag;
                    continue;
                }
                if(LEFT.empty() || (!RIGHT.empty() && LEFT.front().t >= RIGHT.front().t)) {
                    now_time = RIGHT.front().t;
                    while(tn < n && !RIGHT.empty() && RIGHT.front().t <= now_time)
                        tn++, res[RIGHT.front().in] = now_time+t, RIGHT.pop();
                    now_time += t;
                } else {
                    if(!LEFT.empty() || RIGHT.empty() && LEFT.front().t < RIGHT.front().t)
                        now_time = max(LEFT.front().t, now_time)+t;
                    else
                        now_time += t;
                }
            }
            flag = 1-flag;
        }
        for(i = 0; i < m; i++)
            printf("%d\n", res[i]);
        if(testcase)
            puts("");
    }
    return 0;
}
/*
1
1113 1072 11
8631 left
12526 left
15884 left
19388 left
24635 left
30627 right
30627 left
31757 left
35635 right
36133 left
40444 left
*/