CF2A Winner(map+思维)

题目描述：
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 mm ) at the end of the game, than wins the one of them who scored at least mm 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.

输入：
The first line contains an integer number nn ( 1<=n<=10001<=n<=1000 ), nn is the number of rounds played. Then follow nn 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.
输出：
Print the name of the winner.

大意：纸牌游戏，进行n回合，每回合一条 名字+分数 信息的组合，找出游戏结束最高分的玩家，如果最高分有多位，找出第一个达到最高分的玩家

思路：
先把结束时的最高分找出来，然后按照输入顺序去找第一个到达最大分的即可，那么如何找最高分玩家里第一个达到最大分的呢，首先这个玩家的最终分要等于最高分，其次在过程中这个玩家的分数要第一个 >= 最大分，有大于是因为他可能掉分，但是他第一个达到了

#include
using namespace std;
 
typedef unsigned long long ull;
typedef long long ll;
 
const ll maxx = 1e18;
const int N = 1e6+100;
const int ps = 1e4+10;
const double eps = 1e-8;

map<string,int >mp,mps;
int n,k,max1;
string s[1001],ans;
int a[1001];

int main()
{
	cin>>n;
	
	for(int i=1;i<=n;i++)
	{
		cin>>s[i]>>a[i];
		mp[s[i]]+=a[i];
	}//第一遍处理，记录最终状态
	
	for(auto k : mp)
	{
		max1=max(max1,k.second);
	}//找出最大分
	
	for(int i=1;i<=n;i++)
	{
		mps[s[i]]+=a[i];
		if(mp[s[i]]==max1&&mps[s[i]]>=max1)
		{
			ans=s[i];
			break;
		}
	}//第二遍处理，找出分数等于最大分且在过程中第一个 >= 最大分的选手
	
	cout<<ans;

}

反思：
这个题的思路特别巧妙，相当于把处理过程做了两遍，第一遍先找出最高分，第二遍找出最先达到最高分的选手，注意可能会有减分的影响；