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 E: Test

A vocational preference test, unlike an aptitude test, seeks to identify careers that the candidate might find satisfying. Based on the answers to a slew of seemingly inane multiple-choice questions like the one below, the test determines which occupations suit the candidate's personality.

   Which would you rather spend an afternoon doing?

   (a) feeding chickens
   (b) driving a race car
   (c) watching The Simpsons on TV
   (d) suntanning
   (e) building a dog house

Each question asks the candidate to express a preference from among five activities, selected from a common larger set. That is, activities like feeding chickens or suntanning are likely to appear in several different questions.

If a candidate answers A in a question containing A, B, C, D, E as alternatives, this choice indicates a preference for A over each of B, C, D, E. Also, if one answer indicates a preference for X over Y and one or more other answers indicate a preference for Y over Z, the combined set of answers indicates a preference for X over Z.

The candidate may provide contradictory answers; that is, the answers may indicate a preference for X over Y and also for Y over X. These contradictions indicate inconsistency, a personalitiy attribute that may suggest a career in politics or used auto sales.

Given a set of answers to a vocational preference test, you are to partition the activities into the minimal number of sets such that, for each pair within a given set, the answers indicate a contradictory preference.

The input contains several test cases followed by a line containing 0. Each case begins with n, the number of questions in the test. n lines follow, each containing the names of five distinct activities, followed by the candidate's answer - one of the five alternatives. Each activity is named by a single upper case letter.

For each test case, output the sets, one per line. Output the elements of each set in alphabetical order, and output the sets in alphabetical order by their least element. The sets should together contain exactly the set of activities that appear in the input. Leave an empty line between test cases.

Sample Input

4
A B C D E C
F C H I J J
K B H I F I
K C E B J K
0

Output for Sample Input

A
B
C
D
E
F
H
I J K

---

Gordon V. Cormack


題目意思 : 以範例輸入來說,
當選項中有 ABCDE 時, 優先選 C, FCHIJ 時 選 J,
那麼你可以想成 C 優先於 A, ... 等,
最後找出最少的集合數量, 每個集合做決策時, 必須是矛盾的情況
做法 : 做成一張有向圖, 找出 SCC, 也就是矛盾情況


#include <cstdio>
#include <iostream>
#include <set>
#include <vector>
#include <algorithm>
#define maxN 100
using namespace std;
vector<int> g[maxN];
set<string> ans;
int vfind[maxN], findIdx;
int stk[maxN], stkIdx;
int in_stk[maxN], visited[maxN];
int scc(int nd) {
    in_stk[nd] = visited[nd] = 1;
    stk[++stkIdx] = nd;
    vfind[nd] = ++findIdx;
    int mn = vfind[nd];
    for(vector<int>::iterator it = g[nd].begin();
        it != g[nd].end(); it++) {
        if(!visited[*it])
            mn = min(mn, scc(*it));
        if(in_stk[*it])
            mn = min(mn, vfind[*it]);
    }
    if(mn == vfind[nd]) {
        string tmp = "";
        do {
            tmp += (stk[stkIdx]+'A');
            in_stk[stk[stkIdx]] = 0;
        } while(stk[stkIdx--] != nd);
        sort(tmp.begin(), tmp.end());
        ans.insert(tmp);
    }
    return mn;
}
int main() {
    int n, i, j, first = 0;
    char x[4], y[5][4];
    while(scanf("%d", &n) == 1 && n) {
        int app[26] = {};
        for(i = 0; i < 26; i++) {
            g[i].clear();
            visited[i] = in_stk[i] = 0;
        }
        for(i = 0; i < n; i++) {
            for(j = 0; j < 5; j++)
                scanf("%s", y[j]);
            scanf("%s", x);
            app[x[0]-'A'] = 1;
            for(j = 0; j < 5; j++) {
                app[y[j][0]-'A'] = 1;
                if(y[j][0] != x[0]) {
                    g[y[j][0]-'A'].push_back(x[0]-'A');
                }
            }
        }
        ans.clear();
        for(i = 0; i < 26; i++) {
            if(!visited[i] && app[i]) {
                findIdx = stkIdx = 0;
                scc(i);
            }
        }
        if(first)
            puts("");
        first = 1;
        for(set<string>::iterator it = ans.begin();
            it != ans.end(); it++) {
            printf("%c", (*it)[0]);
            for(int j = 1; j < (*it).length(); j++)
                printf(" %c", (*it)[j]);
            puts("");
        }
    }
    return 0;
}