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Description
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 Input
3 mike 3 andrew 5 mike 2
andrew
3 andrew 3 andrew 2 mike 5
andrew
题目大意： 玩一个游戏，每轮有一个人得分，或减分，当所有轮完成后，找出分数最高的那个人，如果是多个人，找出那些人中第一个达到最高分或更高的人。
先保存好每个人的分数，记录下究竟有那几个人会得奖，再次遍历所有回合，找出第一个高于等于那个分数的且最终会得奖的人。
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std ;
map <string,int> Map ;
struct node{
char str[40] ;
int c ;
}p[1100] , q[1100] ;
int c[1100] , num[1100];
int main()
{
int n , i , k , cnt = 1 ;
Map.clear() ;
memset(c,0,sizeof(c)) ;
memset(num,0,sizeof(num)) ;
scanf("%d", &n) ;
for(i = 0 ; i < n ; i++)
{
scanf("%s %d", p[i].str, &p[i].c) ;
k = Map[p[i].str] ;
if( k == 0 )
{
k = cnt ;
strcpy(q[k].str,p[i].str) ;
Map[ p[i].str ] = cnt++ ;
}
c[k] += p[i].c ;
}
int max1 = -2000000 ;
for(i = 0 ; i < cnt ; i++)
{
max1 = max(max1,c[i]) ;
}
for(i = 0 ; i < cnt ; i++)
{
if( c[i] == max1 )
num[i] = 1 ;
}
memset(c,0,sizeof(c)) ;
for(i = 0 ; i < n ; i++)
{
k = Map[ p[i].str ] ;
c[k] += p[i].c ;
if( c[k] >= max1 && num[k] )
break ;
}
printf("%s\n", q[k].str) ;
return 0;
}
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原文地址：http://blog.csdn.net/winddreams/article/details/42671815
