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 D - Australian Voting

Australian ballots require that the voter rank the candidates in order of choice. Initially only the first choices are counted and if one candidate receives more than 50% of the vote, that candidate is elected. If no candidate receives more than 50%, all candidates tied for the lowest number of votes are eliminated. Ballots ranking these candidates first are recounted in favour of their highest ranked candidate who has not been eliminated. This process continues [that is, the lowest candidate is eliminated and each ballot is counted in favour of its ranked non-eliminated candidate] until one candidate receives more than 50% of the vote or until all candidates are tied.

Input

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

The first line of input is an integer n <= 20 indicating the number of candidates. The next n lines consist of the names of the candidates in order. Names may be up to 80 characters in length and may contain any printable characters. Up to 1000 lines follow; each contains the contents of a ballot. That is, each contains the numbers from 1 to n in some order. The first number indicates the candidate of first choice; the second number indicates candidate of second choice, and so on.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

The Output consists of either a single line containing the name of the winner or several lines containing the names of the candidates who tied.

Sample Input

1

3
John Doe
Jane Smith
Sirhan Sirhan
1 2 3
2 1 3
2 3 1
1 2 3
3 1 2

Output for Sample Input

John Doe

---

#include <stdio.h>
#include <iostream>
#include <sstream>
using namespace std;
int main() {
    int t,n, i, j;
    scanf("%d", &t);
    getchar();
    while(t--) {
        scanf("%d", &n);
        getchar();
        string name[25], line;
        int vote[1005][25];
        for(i = 1; i <= n; i++)
            getline(cin, name[i]);
        int m = 0;
        while(getline(cin, line)) {
            if(line == "")  break;
            stringstream sin;
            sin << line;
            int idx = 0;
            while(sin >> vote[m][idx])
                idx++;
            m++;
        }
        int Idx[1005] = {}, fire[25] = {};
        while(1) {
            int p[25] = {}, tot = 0;
            for(i = 0; i < m; i++) {
                while(Idx[i] < n && fire[vote[i][Idx[i]]])
                    Idx[i]++;
                if(Idx[i] < n) {
                    p[vote[i][Idx[i]]]++;
                    tot++;
                }
            }
            int half = tot/2 + tot%2;
            int mx = 0, mi = 0xffff;
            for(i = 1; i <= n; i++) {
                if(!fire[i]) {
                    if(p[i] > mx)
                        mx = p[i];
                    if(p[i] < mi)
                        mi = p[i];
                }
            }
            if(mx == mi || mx >= half) {
                for(i = 1; i <= n; i++)
                    if(p[i] == mx)
                        cout << name[i] << endl;
                break;
            }
            for(i = 1; i <= n; i++)
                if(p[i] == mi)
                    fire[i] = 1;
        }
        if(t)
            puts("");
    }
    return 0;
}