【codeforces#2A】Winner

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.
Sample test(s)
Input

3
mike 3
andrew 5
mike 2

Output

andrew

Input

3
andrew 3
andrew 2
mike 5

Output

andrew

题意：玩游戏，每轮有胜出或失败的人的名字，以及得到或失去的分数，判断最后谁的分数最高，如果最高分有多个人，输出这些人中最先达到这个分数的名字。

解题思路：用map将所有人得到的最终分数存好，然后得出最大的分数，在是最大的分数的这些人中选出最先得到这个分数的人，输出。

code：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
using namespace std;
struct round
{
    char name[35];
    int scr;
}r[1005];
int main()
{
    int n,i;
    map<string,int>r1,r2;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%s%d",r[i].name,&r[i].scr);
        r1[r[i].name]+=r[i].scr;
    }

    int mx=-1000005;
    for(i=1;i<=n;i++){
            if(r1[r[i].name]>mx){
                mx=r1[r[i].name];
            }
    }

    for(i=1;i<=n;i++){
        r2[r[i].name]+=r[i].scr;
        if(mx==r1[r[i].name]&& r2[r[i].name]>=mx){
            printf("%s\n",r[i].name);
            break;
        }

    }

    return 0;
}