Codeforces 2A Winner （STL map使用）

A. Winner

time limit per test
1 second

memory limit per test
64 megabytes

input
standard input

output
standard output

The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number
of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name
score", where name is a player's name, and score is
the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to m)
at the end of the game, than wins the one of them who scored at least m points first. Initially
each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.

Input

The first line contains an integer number n (1  ≤  n  ≤  1000), n is
the number of rounds played. Then follow n lines, containing the information about the rounds in "name
score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is
an integer number between -1000 and 1000, inclusive.

Output

Print the name of the winner.

题意：很多人玩一个游戏，每一轮有一个人得分或者扣分，最后分数最高的人夺冠；如果最后有多个人分数都是最高的，则这些人里面，在比赛过程中首先达到或者超过这个分数的人夺冠。现在给定最多1000轮每轮的情况，求最后的冠军是谁。

用map做，也是简单的不行。。

#include <iostream>
#include <map>
#include <string.h>
#include <algorithm>
using namespace std;
int score[1010];
map<string, int> p1,p2;
string name[1010];
int main()
{
int n,i,j;
while(cin>>n)
{
int m = 0;
for(i = 0;i < n;i ++)
{
cin>>name[i]>>score[i];
p1[name[i]] += score[i];
}
for(i = 0;i < n;i ++)
if(p1[name[i]] > m) m = p1[name[i]];
for(i = 0;i < n;i ++)
{
p2[name[i]] += score[i];
if(p2[name[i]] >= m&&p1[name[i]] == m)
{
cout<<name[i]<<endl;
break;
}
}
}
}