【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

題意：玩遊戲，每輪有勝出或失敗的人的名字，以及得到或失去的分數，判斷最後誰的分數最高，如果最高分有多個人，輸出這些人中最先達到這個分數的名字。

解題思路：用map將所有人得到的最終分數存好，然後得出最大的分數，在是最大的分數的這些人中選出最先得到這個分數的人，輸出。

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;
}