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 J
Setting Problems

Input: Standard Input

Output: Standard Output

As you see, setting problems fora programming contest is a tough job. There are so many things to do likecreating problems, solutions, data files, verification of problem statements,writing alternate judge solutions etc. etc. Given the responsibility ofcreating the problemset for ‘If you can not crash judges by solutions, crashcontestants by problems’ programming contest, Sultan & GolapiBaba haverealized that to the backbone. Finally they agree that they will set Nproblems for the contest. For each of the problems, first Sultan will createthe problem statement, solution & i/o data. Afterhe finishes his work, GolapiBaba does the verification & alternate solutionwriting part for that particular problem. Each of them needs a particularamount of time to complete their tasks for a certain problem. Also, none ofthem works on more than one problem at a time. Note that, GolapiBaba can startworking on a problem immediately after Sultan finishes that problem or he maywish to start that problem later.

You will be given the times that Sultan& GolapiBaba requires to complete their respective tasks for every singleproblem. Determine the minimum possible time required to complete the wholeproblemset.

Input

There are around 50 test cases. Eachtest case starts with a single integer N (1<=N<=20), thenumber of problems in the contest. The next line contains N integers S_i(1<=S_i<=100, 1<=i<=N) where Sidenotes the time required for Sultan to complete his tasks for problem i. The nextline has N more integers G_i (1<=G_i<=100,1<=i<=N) where G_i denotes the time required forGolapibaba to complete his tasks on problem i.

Output

For each test case, print theminimum time required to complete the problemset.

Problemsetter: Mohammad Mahmudur Rahman

Special Thanks To: Abdullah Al Mahmud

題目描述：

一個人先把題目生出來，然後緊接另一個人進行檢查動作。每一題分別有兩個時間消耗，對於生題目以及檢查的時間，如何排列順序，使得時間花費長度最短。兩個人是同時進行動作的，但第二個人總是要等題目創完才能進行該題的檢查。

題目解法：

根據兩個相鄰的方案進行排序，又或者可以參照　

http://www3.tcgs.tc.edu.tw/npsc/index.php?topic=118.0

greedy 就是了。

#include <stdio.h>
#include <algorithm>
using namespace std;
struct E {
    int s, q;
    bool operator<(const E &a) const {
        return s+max(a.s, q)+a.q < a.s+max(s, a.q)+q;
    }
};
int main() {
    int n, i;
    E D[105];
    while(scanf("%d", &n) == 1) {
        for(i = 0; i < n; i++)
            scanf("%d", &D[i].s);
        for(i = 0; i < n; i++)
            scanf("%d", &D[i].q);
        sort(D, D+n);
        int ret = 0, ss = 0;
        for(i = 0; i < n; i++) {
            ss += D[i].s;
            ret = max(ret, ss);
            ret = ret+D[i].q;
        }
        printf("%d\n", ret);
    }
    return 0;
}