hdu 2648 Shopping

Shopping
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1400 Accepted Submission(s): 456


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


Recommend
yifenfei

//1218MS    1136K    754 B    C++
//开始还想用排序，后来想想只用容器就够了
//代码好像也写复杂了，id不用另外记录
//在循环中每次循环比较一次就行了 
#include<iostream>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
struct node{
    int id;
    int s;
}a[10005];
int main(void)
{
    int n,m,p,id;
    string c;
    map<string,int>M;
    while(cin>>n)
    {
        M.clear();
        for(int i=0;i<n;i++){
            cin>>c;
            M[c]=i;
            if(c=="memory") id=i;
            a[i].s=0;
            a[i].id=i;
        }
        cin>>m;
        while(m--){
            for(int i=0;i<n;i++){
                cin>>p>>c;
                a[M[c]].s+=p;
            }
            int cnt=0;
            for(int i=0;i<n;i++)
                if(a[i].s>a[id].s) cnt++;
            printf("%d\n",cnt+1);
        }
    }
    return 0;
}