1.6.8 Australia votes Australian voting

Pc/uva 110108/10142

Simulation.

AUTHOR:CHC//first Edit time:2014-01-14 14:37//last edit time:2014-01-15 22:03//filename:1.cpp #include <iostrea m> #include <cstdio> #include <string.h> #include <queue> #include <vector> #include <
Algorithm> using namespace std;
int N,t,ns;
Char s[30][100];
Char ts[1000];
Vector <int> vote[1010];
Vector <int>:: iterator it;
int ha[30];
    void getnum (int kn) {int i=0;
        while (Ts[i]) {while (ts[i]== ' &&ts[i]) i++;
        int tt=0;
            while (ts[i]!= ' &&ts[i]) {tt=tt*10+ts[i]-' 0 ';
        ++i;
    } if (tt!=0) Vote[kn].push_back (TT);
    }} void voted () {for (int i=0;i<ns;i++) {ha[vote[i][0]]++;
        } while (1) {int ping=1,min=1010,first=-1;
                for (int i=0;i<n;i++) {if (HA[I+1]&GT;NS/2) {printf ("%s\n", S[i]);
            return; }/* if (ha[i+1]>=0&&first==-1) FIRST=HA[I+1];
            if (First>=0&&ha[i+1]>=0&&ha[i+1]!=first) ping=0;
            if (ha[i+1]>=0&&ha[i+1]<min) min=ha[i+1];
                *////* if (ha[i+1]>=0) {if (min!=1010&&min!=ha[i+1]) ping=0;
            if (min>ha[i+1]) min=ha[i+1];
                    }//*/} if (ping) {for (int i=0;i<n;i++) if (ha[i+1]>=0)
            printf ("%s\n", S[i]);
        return;
            } for (int i=0;i<ns;i++) {int flag= (min==ha[vote[i][0]]);
                    For (It=vote[i].begin (); It!=vote[i].end (); it++) if (ha[*it]==min) {vote[i].erase (IT);
                it--;
        } if (flag) ha[vote[i][0]]++;
    } for (int i=0;i<n;i++) if (ha[i+1]==min) ha[i+1]=-1;
    }} int main () {scanf ("%d", &t);
        while (t--) {scanf ("%d", &n);

        GetChar ();
        for (int i=0;i<1010;i++) vote[i].clear ();

        memset (ha,0,sizeof (ha));
        for (int i=0;i<n;i++) {gets (s[i]);
        } ns=0;
            while (gets (TS)!=null&&strcmp (ts, "")!=0) {getnum (NS);
        ++ns;
        } voted ();
    if (t!=0) puts ("");
} return 0;
 }

Australian voting Australian ballots require that voters rank all the candidates in order of choice. Initially only the first choices is counted, and if one candidate receives more than 50% of the vote then that candidate Is elected. However, if no candidate receives more than 50%, all candidates tied for the lowest number of votes is eliminated. Ballots ranking these candidates first is recounted in favor of their highest-ranked non-eliminated candidate. This process of eliminating the weakest candidates and counting their ballots in favor of the preferred non-eliminated can Didate continues until one candidate receives more than 50% of the vote, or until all remaining candidates is tied. Input The input begins with a single positive integer on a line by itself indicating the number of cases following, each as Desc Ribed below. This was followed by a blank line. There is also a blank line between the consecutive inputs. The first line of all case are an integer N20 indicating the number of candidates. The next n lines consist of the names of the candidates in order, each up to a characters in length and containing any PR Intable characters. Up to $ lines follow, each containing the contents of a ballot. Each ballot 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 The output of each test case consists of either a single line containing the name of the winner or several lines Containin G The names of all candidates who is tied. The output of each of the consecutive cases is separated by a blank line. Sample Input 1 3 John Doe Jane Smith Jane Austen 1 2 3 2 1 3 2 3 1 1 2 3 3 1 2 Sample Output John Doe