POJ 2153 排名次 水题map

Rank List

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 8067		Accepted: 2629

Description
Li Ming is a good student. He always asks the teacher about his rank in his class after every exam, which makes the teacher very tired. So the teacher gives him the scores of all the student in his class and asked him to get his
rank by himself. However, he has so many classmates, and he can’t know his rank easily. So he tends to you for help, can you help him?
Input
The first line of the input contains an integer N (1 <= N <= 10000), which represents the number of student in Li Ming’s class. Then come N lines. Each line contains a name, which has no more than 30 letters. These names represent
all the students in Li Ming’s class and you can assume that the names are different from each other.

In (N+2)-th line, you'll get an integer M (1 <= M <= 50), which represents the number of exams. The following M parts each represent an exam. Each exam has N lines. In each line, there is a positive integer S, which is no more then 100, and a name P, which
must occur in the name list described above. It means that in this exam student P gains S scores. It’s confirmed that all the names in the name list will appear in an exam.

Output
The output contains M lines. In the i-th line, you should give the rank of Li Ming after the i-th exam. The rank is decided by the total scores. If Li Ming has the same score with others, he will always in front of others in the
rank list.
Sample Input

3
Li Ming
A
B
2
49 Li Ming
49 A
48 B
80 A
85 B
83 Li Ming

Sample Output

1
2

Source
POJ Monthly,Li Haoyuan

题意：

输入n 表示有n个人 然后输入n个人名   在输入m  表示m次考试 每次考试成绩是累加的  对于每次考试都要输出liming的名次

思路：

map  注意map是按照前面的元素排序的 从小到大

#include<stdio.h>
#include<string>
#include<map>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
int i,j,k,n,m,cnt;
char s[100];
while(scanf("%d",&n)!=EOF)
{
map<string,int>mp;
map<string,int>::iterator it,it2;
getchar();
for(i=0;i<n;i++)
{
gets(s);
mp[s]=0;
}
scanf("%d",&m);
for(i=0;i<m;i++)
{
cnt=0;
for(j=0;j<n;j++)
{
scanf("%d",&k);
getchar();
gets(s);
mp[s]+=k;
}
for(it=mp.begin();it!=mp.end();it++)
{
if(it->second>mp["Li Ming"]) cnt++;
}
printf("%d\n",cnt+1);
}
}
return 0;
}