poj2643 Election (map)

题目：http://poj.org/problem?id=2643

Output

Output consists of a single line containing one of: 

• The name of the party with whom the winning candidate is associated, if there is a winning candidate and that candidate is associated with a party. 
  
• The word "independent" if there is a winning candidate and that candidate is not associated with a party. 
  
• The word "tie" if there is no winner; that is, if no candidate receives more votes than every other candidate.

Sample Input

3Marilyn MansonRhinocerosJane DoeFamily CoalitionJohn Smithindependent6John SmithMarilyn MansonMarilyn MansonJane DoeJohn SmithMarilyn Manson

Sample Output

Rhinoceros

题意：

有n个候选人，每个人有两行数据，第一行为姓名 ，第二行为所属党派。

有m张选票，每张一行，代表所投的候选人的姓名，若所投的姓名不在上述名单中，应忽略。

若有唯一的人胜选，输出其所属党派，若多人同时获最高票，输出“tie”.

#include<stdio.h>#include<string.h>#include<map>#include<iostream>using namespace std;int main(){    int n,m;    char name[100];    char party[100];    scanf("%d",&n);    getchar();    map<string ,string > name1;    map<string ,int > name2;    while(n--)    {      gets(name);      gets(party);       name1[name]=party;    }    scanf("%d",&m);    getchar();    int s=0;    char ans[100];    int flag=1;    while(m--)    {        gets(name);        if(name1[name][0]!='\0')name2[name]++;  //判断被投票人的姓名是否在名单中        if(name2[name]>s)        {            s=name2[name];            strcpy(ans,name);            flag=1;        }        else            if(name2[name]==s)flag=0;    }    if(flag)printf("%s\n",name1[ans].c_str());    else printf("tie\n");    return 0;}