链接：click here
题意：
A Simple Problem
时间限制：3000ms | 内存限制：65535KB
难度：2
描述You know, just as the title imply, this is a simple problem. In a contest, given the team-id, solved, penalty of all the teams, tell me the champion.If the numbers of solved problem of two team are different, the rank of the one who solves more problems is higher. Otherwise, if the the penalties of two team are different, the rank of the one who has less penalty is higher. Otherwise, the rank of the one whose team-id’s lexicographic order is earlier than the other is higher.输入The first line of the input is an integer T which stands for the number of test cases. Then T test cases follow.The first line of test case is a number n, which is the number of team in a contest. Then n line(s) follow. Each line contain a string, and two integers: str, s( 0 <= s <= 15 ), p(0<=p <= 20000), separated by a blank indicating that there is a team whose id is str,the number of solved problem is s, and the penalty is p.constraints:n is in the range of [1 100].Each team-id does not contain any blanks.The length of team-id is in the range of[1, 20].Any two teams will not have the same team-id.输出For each test case, output one line with an string indicating the the champion.样例输入17Refreshing 5 745 Rock_Restart 4 510 LeadWill 4 679 APTX4869 5 374 WaterCop 5 607 ISAP 5 638 TLE 4 902 样例输出APTX4869来源SCU Programming Contest 2011 Preliminary就是按照solved problem，（越大排前），penalties（越小排前），，id（字典序排名）的规则排序：
代码：
#include <math.h>#include <queue>#include <deque>#include <vector>#include <stack>#include <stdio.h>#include <ctype.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>using namespace std;#define Max(a,b) a>b?a:b#define Min(a,b) a>b?b:a#define mem(a,b) memset(a,b,sizeof(a))int dir[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};const double eps = 1e-6;const double Pi = acos(-1.0);static const int inf= ~0U>>2;static const int maxn =110;struct node{char id[25];int num;int time;}aa[maxn];bool cmp(node a,node b){return a.num!=b.num?a.num>b.num:a.time!=b.time?a.time<b.time:strcmp(a.id,b.id)<0;}int main(){int t,n,i;scanf("%d",&t);while(t–){scanf("%d",&n);for(i=0; i<n; i++)scanf("%s%d%d",aa[i].id,&aa[i].num,&aa[i].time);sort(aa,aa+n,cmp);//printf("%d\n",aa[0].num);printf("%s\n",aa[0].id);}return 0;} When you want to give up, think of why you persist until now!
躲在墙角、掩藏那孤独而又不奢怜悯的伤…
