Sicily 1299 Academy Awards （map + vector）集装箱

链接：http://soj.me/show_problem.php?pid=1299&cid=

Description

Selected from 3,850 teams from 1,329 universities in 68 countries competing at 106 sites and preliminary contests worldwide, sixty-eight teams competed for bragging rights and prizes at The 27th Annual ACM International Collegiate Programming Contest World Finals sponsored by IBM on March 25, 2003, in Hollywood, California. The 2003 World Champion is Warsaw University . And Zhongshan University won the 8^th place. During those days, another world famous event was held in the same place. It was the 75^th Annual Academy Awards. It’s also known as Oscar.

   We always say that the Best Picture is the most important award of all the awards. Before the Oscar Night, we can’t tell which film will win Best Picture. Fortunately, we can dope it out from the Nominee List of all the awards other than the Best Picture. I suggest that you should follow my 3 rules here. 

l         All the films in the list have chances to win the Best Picture

l         The film which will win the Best Picture is the film which has been nominated the most times in the list

l         If there are more than one film which have been nominated the most times in the list, we will choose the first one which appears in the list

 Let’s see such a List below.  

VISUAL EFFECTS

 THE LORD OF THE RINGS: THE TWO TOWERS

SPIDER-MAN

STAR WARS EPISODE II ATTACK OF THE CLONES

 SOUND EDITING  

THE LORD OF THE RINGS: THE TWO TOWERS

MINORITY REPORT

ROAD TO PERDITION

       From the list, we can find that THE LORD OF THE RINGS: THE TWO TOWERS has been nominated twice. And each of the other films has been nominated only once. So we can say THE LORD OF THE RINGS: THE TWO TOWERS will win the Best Picture. 

Your task is to write a program to figure out the anticipatory winner from the list.

Input

The input file will consist of several lists. The first line of each list contains only one integer n (1≤n≤100), representing the number of awards in the list. Then you get n blocks. Each block indicated the nominees of a distinct award. The first line of each block is the name of the award which is not longer than 80. The second line is m_i (1≤m_i≤10, 1≤i≤n) - the number of nominated films. In the following lines are m_i film names, one per line. For make the question simple, you can assume that there isn’t any space in the film names.

 

The input is terminated by a line with one zero.

Output

For each list, you are supposed to figure out the winner of Best Picture in a single line.

Sample Input

 Copy sample input to clipboard

2VISUAL_EFFECTS3THE_LORD_OF_THE_RINGS:_THE_TWO_TOWERSSPIDER-MANSTAR_WARS_EPISODE_II_ATTACK_OF_THE_CLONESSOUND_EDITING3THE_LORD_OF_THE_RINGS:_THE_TWO_TOWERSMINORITY_REPORTROAD_TO_PERDITION0

Sample Output

THE_LORD_OF_THE_RINGS:_THE_TWO_TOWERS

分析： 统计出现频率最高的电影名字。看一下数据范围都不大，时间复杂度为O(n*m), 统计频率用map容器就能够搞定，可是另一个重要的问题就是，题目中要求电影的顺序，所以没办法啦，加一个vector 就OK啦。

。。

代码例如以下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <vector>
#include <string>
#include <iterator>
#define RST(N)memset(N, 0, sizeof(N))
using namespace std;

vector <string> v;    
map <string, int> mp;
map <string, int>:: iterator it;
string award, file, flag;   //奖项名字，电影名字，出现频率最高的电影名字；
int n, m;   //奖项数，获得每一个奖项的电影名字总数；

int main()
{
    while(cin >> n && n) {
        mp.clear(), v.clear();   //初始化容器。
        while(n--) {
            cin >> award;
            cin >> m;
            for(int i=0; i<m; i++) {
                cin >> file;
                v.push_back(file);
                it = mp.find(file);    //找到该电影出如今map容器中的位置；
                if(it != mp.end()) mp[file]++;  //找到，频率加1；
                else mp[file] = 1;   //未找到，频率为1；并加到map容器其中；
            }
        }
        int max = 0;
        for(int i=0; i<v.size(); i++) {   //找出出现频率最高的电影名字；
            it = mp.find(v[i]);
            if(it->second > max) {
                max = it->second;
                flag = it->first;
            }
        }
        cout << flag << endl;
    }
    return 0;
}

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