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

The Argentine football team coach, the great Diego Maradona, is going to try out a new formation this year. Formation describes how the players are positioned on the pitch.  Instead of the conventional 4-4-2 or 4-3-3, he has opted for 5-5. This means there are 5 attackers and 5 defenders.

You have been hired by Argentina Football Federation (AFF) to write a code that will help them figure out which players should take the attacking/defensive positions.

Maradona has given you a list containing the names of the 10 players who will take the field. The attacking ability and the defensive ability of each player are also given. Your job is to figure out which 5 players should take the attacking positions and which 5 should take the defensive positions.

The rules that need to be followed to make the decision are:

• The sum of the attacking abilities of the 5 attackers needs to be maximized
• If there is more than one combination, maximize the sum of the defending abilities of the 5 defenders
• If there is still more than one combination, pick the attackers that come lexicographically earliest.

Input

The first line of input contains an integer T(T<50) that indicates the number of test cases. Each case contains exactly 10 lines. The i^th line contains the name of the i^th player followed by the attacking and defending ability of that player respectively. The length of a player’s name is at most 20 and consists of lowercase letters only. The attacking/defending abilities are integers in the range [0, 99].

Output

The output of each case contains three lines. The first line is the case number starting from 1. The next line contains the name of the 5 attackers in the format (A₁, A₂, A₃, A₄, A₅) where A_i is the name of an attacker. The next line contains the name of the 5 defenders in the same format. The attackers and defenders names should be printed in lexicographically ascending order. Look at the sample for more details.

Problemsetter: Sohel Hafiz, Special Thanks: Shamim Hafiz, Jane Alam jan

先排序，之後就不用考慮名字的字典順序問題，按照字典順序窮舉就行了

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
struct p {
    string name;
    int a, b;
};
p P[10];
int at, def;
bool cmp(p a, p b) {
    int i;
    for(i = 0; i < a.name.length() && i < b.name.length(); i++)
        if(a.name[i] > b.name[i])
            return 0;
        else if(a.name[i] < b.name[i])
            return 1;
    return a.name.length() < b.name.length();
}
int A[10], B[10];
int aA[10], aB[10];
void dfs(int idx, int a, int d, int ta, int td) {
    if(idx == 10) {
        if(ta > at || (ta == at && td > def)) {
            at = ta, def = td;
            int i;
            for(i = 0; i < 5; i++)
                aA[i] = A[i], aB[i] = B[i];
        }
        return;
    }
    if(a != 5) {
        A[a] = idx;
        dfs(idx+1, a+1, d, ta+P[idx].a, td);
    }
    if(d != 5) {
        B[d] = idx;
        dfs(idx+1, a, d+1, ta, td+P[idx].b);
    }
}
int main() {
    int t, cases = 0, i;
    scanf("%d", &t);
    while(t--) {
        for(i = 0; i < 10; i++)
            cin >> P[i].name >> P[i].a >> P[i].b;
        sort(P, P+10, cmp);
        at = 0, def = 0;
        dfs(0, 0, 0, 0, 0);
        printf("Case %d:\n", ++cases);
        printf("(");
        for(i = 0; i < 5; i++) {
            if(i)   printf(", ");
            cout << P[aA[i]].name;
        }
        printf(")\n");
        printf("(");
        for(i = 0; i < 5; i++) {
            if(i)   printf(", ");
            cout << P[aB[i]].name;
        }
        printf(")\n");
    }
    return 0;
}