HDU 2648 Shopping（map）

Shopping

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2572    Accepted Submission(s): 893

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

曾是驚鴻照影來

 

題意：有N個店，他們的商品價格每天都在上漲，問你 ith天有個叫memory的店，它的價格在所有商店中，有幾個高過它，

            輸出它的排名，有k個高過它，它就是第k+1名。

map水過。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <string>
#define  N 100005

using namespace std;
int n;
map<string,int>mp;
char s[100];

int main() {
   // freopen("test.in","r",stdin);
    while(~scanf("%d",&n)) {
        mp.clear();
        for(int i=1; i<=n; i++) {
            scanf("%s",s);
            mp[s]=0;
        }
        map<string,int>::iterator it;
        int d;
        scanf("%d",&d);
        while(d--) {
            int x;
            for(int i=1; i<=n; i++) {
                scanf("%d%s",&x,s);
                mp[s]+=x;
            }
            int k=0;
            for(it=mp.begin(); it!=mp.end(); it++) {
                if(it->second>mp["memory"]) {
                    k++;
                }
            }
            printf("%d\n",++k);
        }
    }
}