map 加hash hdu 2648 Shopping

Shopping

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3447 Accepted Submission(s): 1277

Problem Description
Every girl likes shopping,so does dandelion.Now she finds the shop is increasing the price every day because the Spring Festival is coming .She is fond of a shop which is called “memory”. Now she wants to know the rank of this shop’s price after the change of everyday.

Input
One line contians a number n ( n<=10000),stands for the number of shops.
Then n lines ,each line contains a string (the length is short than 31 and only contains lowercase letters and capital letters.)stands for the name of the shop.
Then a line contians a number m (1<=m<=50),stands for the days .
Then m parts , every parts contians n lines , each line contians a number s and a string p ,stands for this day ,the shop p ‘s price has increased s.

Output
Contains m lines ,In the ith line print a number of the shop “memory” ‘s rank after the ith day. We define the rank as :If there are t shops’ price is higher than the “memory” , than its rank is t+1.

Sample Input
3
memory
kfc
wind
2
49 memory
49 kfc
48 wind
80 kfc
85 wind
83 memory

Sample Output
1
2

Author
dandelion

Source
曾是惊鸿照影来

题意：
给出m个商店，然后再给出n组测试数据，每组数据有一个sum，一个字符串，判断memory的累加排名；
思路：
这个题用map就可以了；
代码：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<map>const int N=1e4;using namespace std;map<string,int>p;string ans="memory";string s[N+5];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0; i<n; i++)        {            cin>>s[i];        }        int m;        p.clear();        cin>>m;        while(m--)        {            int num=0;            for(int i=0; i<n; i++)            {                int t;                string str;                cin>>t>>str;                p[str]+=t;            }            for(int i=0;i<n;i++)            {                if(p[s[i]]>p[ans])                {                    num++;                }            }            cout<<num+1<<endl;        }    }}