hdu Let the Balloon Rise

最新推荐文章于 2021-06-24 22:53:24 发布

风吹过的烟花

最新推荐文章于 2021-06-24 22:53:24 发布

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分类专栏： ACM算法_STL 文章标签： map

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78550410

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本文介绍了一道编程比赛题目，通过使用C++实现的map数据结构来统计最多出现的颜色。文章提供了完整的源代码，并解释了如何利用迭代器遍历map以找到出现次数最多的气球颜色。

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Let the Balloon Rise

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) : 1

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Problem Description

Input

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

Output

For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Sample Input

5
green
red
blue
red
red
3
pink
orange
pink
0

Sample Output

red
pink

利用了map的特性，然后便利map，用迭代器

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<string>
using namespace std;
int main()
{
    int n;
    char a[20];
    while(scanf("%d",&n),n)
    {
        map<string,int>words;
        for(int i=0; i<n; i++)
        {
            cin>>a;
            words[a]++;
        }
        int k=0;
        map<string,int>::iterator it;
        it=words.begin();
        while(it!=words.end())
        {
            if(it->second>k)
            {
                k=it->second;
            }
            it++;
        }
        it=words.begin();
        while(it!=words.end())
        {
            if(it->second==k)
            {
                cout<<it->first<<endl;;
                break;
            }
            it++;
        }
    }
    return 0;
}

map的迭代

定义一个迭代器（以int&int的map为例）：map<int,int>::iterator mapi;
然后遍历map就可以写成
for(mapi=map.begin;mapi!=map.end;mapi++)
{
    int a=mapi->first;

    int b=mapi->second;

}