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 A: Ferry Loading IV

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 an l-meter-long ferry that crosses the river. 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 as 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 cars that are waiting to cross as long as they fit on its deck. The cars are loaded in the order of their arrival; ferry's deck accommodates only one lane of cars. The ferry is initially on the left bank where it broke and it took quite some time to fix it. In the meantime, lines of cars formed on both banks that await to cross the river.

The first line of input contains c, the number of test cases. Each test case begins with l, m. m lines follow describing the cars that arrive in this order to be transported. Each line gives the length of a car (in centimeters), and the bank at which the car arrives ("left" or "right").

For each test case, output one line giving the number of times the ferry has to cross the river in order to serve all waiting cars.

Sample input

4
20 4
380 left
720 left
1340 right
1040 left
15 4 
380 left
720 left
1340 right
1040 left
15 4 
380 left
720 left
1340 left
1040 left
15 4 
380 right
720 right
1340 right
1040 right

Output for sample input

3
3
5
6

---

Piotr Rudnicki

#include <stdio.h>
#include <queue>
using namespace std;

int main() {
    int t, n, m;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        n = n*100;
        queue<int> L, R;
        int l;
        char s[50];
        while(m--) {
            scanf("%d %s", &l, s);
            if(s[0] == 'l')
                L.push(l);
            else
                R.push(l);
        }
        int flag = 0, res = 0;
        while(!L.empty() || !R.empty()) {
            res++;
            int sum = 0;
            if(flag) { // right
                while(!R.empty() && sum + R.front() <= n)
                    sum += R.front(), R.pop();
            } else {
                while(!L.empty() && sum + L.front() <= n)
                    sum += L.front(), L.pop();
            }
            flag = 1-flag;
        }
        printf("%d\n", res);
    }
    return 0;
}