CUC-SUMMER-CONTEST-1-C

C - Winner

CodeForces-2A

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.

Example
Input
3
mike 3
andrew 5
mike 2
Output
andrew
Input
3
andrew 3
andrew 2
mike 5
Output
andrew

题意：算参赛选手的总分，分最高的获胜，如果不止一个人最高分，那么谁最先达到此分数，谁获胜。

解法：用map存，分两次比较，第一次找出最高分，第二次找到最先达到最高分的选手。

代码：

#include<iostream>
#include<map>
#include<string>
using namespace std;
int score[1005];
string name[1005];
int main()
{
    int n,maxn=0;
    cin>>n;
    map<string,int> a;
    map<string,int> b;
    string ans;
    for(int i=0;i<n;i++){
        cin>>name[i];
        cin>>score[i];
        a[name[i]]+=score[i];
    }
    for(int i=0;i<n;i++)
        if(maxn<a[name[i]])
            maxn=a[name[i]];
    for(int i=0;i<n;i++){
        if(a[name[i]]==maxn){
            b[name[i]]+=score[i];
            if(b[name[i]]>=maxn){
                ans=name[i];
                break;
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}