zoukankan      html  css  js  c++  java
  • hdu 1004 Let the Balloon Rise strcmp、map、trie树

    Let the Balloon Rise

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

    Problem Description
    Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

    This year, they decide to leave this lovely job to you. 
     
    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
     
    Author
    WU, Jiazhi
    题意:找出出现次数最多的那个字符串;
    strcmp:
    #include<stdio.h>
    #include<string.h>
    char a[1000][20];
    int b[1011];
    int main ()
    {
        int x,y,z,j,k,i,t,max;
        while(scanf("%d",&x)!=EOF)
        {
            if(x==0)
            break;
            getchar();
            memset(b,0,sizeof(b));
            for(i=0;i<x;i++)
            scanf("%s",a[i]);
            for(i=0;i<x;i++)
            for(t=0;t<x;t++)
            if(strcmp(a[i],a[t])==0)
            b[i]++;
            max=b[0];
            k=0;
            for(i=1;i<x;i++)
            if(max<b[i])
            {max=b[i];k=i;}
            printf("%s
    ",a[k]);
        }
        return 0;
    }

    map:

    #include<stdio.h>
    #include<map>
    #include<iostream>
    #include<string.h>
    #include<string>
    using namespace std;
    int main()
    {
        int x,y,z,i,t,max;
        string a,b;
        map<string,int>p;
        while(scanf("%d",&x)!=EOF)
        {
            getchar();
            p.clear();
            if(x==0) break;
            max=0;
            for(i=0;i<x;i++)
            {
                cin>>a;
                p[a]++;
                if(p[a]>max)
                {
                    max=p[a];
                b=a;}
            }
            cout<<b<<endl;
        }
        return 0;
    }

    trie树:

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define ll long long
    #define mod 1000000007
    #define pi (4*atan(1.0))
    const int N=1e5+10,M=5e6+10,inf=1e9+10;
    int a[N][27],sum[M],len,ans=0;
    void init()
    {
        memset(a,0,sizeof(a));
        memset(sum,0,sizeof(sum));
        len=1;
        ans=0;
    }
    char aaa[N];
    int getnum(char a)
    {
        return a-'a';
    }
    void insertt(char *aa)
    {
        int u=0,n=strlen(aa);
        for(int i=0; i<n; i++)
        {
            int num=getnum(aa[i]);
            if(!a[u][num])
            {
                a[u][num]=len++;
            }
            u=a[u][num];
            sum[u]++;
        }
        if(sum[u]>ans)
        {
            ans=sum[u];
            strcpy(aaa,aa);
        }
    }
    int getans(char *aa)
    {
        int u=0,x=strlen(aa);
        for(int i=0; i<x; i++)
        {
            int num=getnum(aa[i]);
            if(!a[u][num])
                return 0;
            u=a[u][num];
        }
        return sum[u];
    }
    char ch[N];
    int main()
    {
        int x,y,z,i,t;
        while(~scanf("%d",&x))
        {
            if(x==0)break;
            init();
            for(i=0; i<x; i++)
            {
                scanf("%s",ch);
                insertt(ch);
            }
            printf("%s
    ",aaa);
        }
        return 0;
    }
     
  • 相关阅读:
    学习python报错处理
    Java命名规则
    python打开文件可以有多种模式
    python中的异常
    python中的构造函数和构造函数和析构函数的作用
    学习网址
    使用Python+selenium过程中所需安装的库和软件
    oracle 、server和my sql 语法区别
    blob(斑点)特征,SimpleBlobDetector(OpenCV案例源码detect_blob.cpp解读)
    查找与绘制轮廓findContours()、drawContours()(OpenCV案例源码contours2.cpp解读)
  • 原文地址:https://www.cnblogs.com/jhz033/p/5581141.html
Copyright © 2011-2022 走看看