UVa - 12450 - SpaceRecon Tournament

先上题目：

Problem G: SpaceRecon Tournament

SpaceRecon, the hottest game of 2011, is a real-time strategy thriller where players control your armies to destroy their opponent. Players may choose from one of the three available races -- Xurks (biological), Protoast (technological), and Earthians (humanoids) -- to build their respective armies by collecting resources from the land and spending them on army units, upgrades, and infrastructure. The player who destroys or outlasts their opponent is victorious.

The game has an intricate single-player story mode where players recreate scenes of the survival of Earthian Commander John Rainard's travels through the Xurkling planet, and also the charades of once-Earthian-now-Xurkling Queen Stephanie Karpenter. After completing the 15 hours of single player gameplay, most users try their hand at multiplayer head-to-head battles online using Actionweb, the number one SpaceRecon game matching hub.

Actionweb hosts tournaments of 2^M players and publishes the results of each tournament by listing each player handle and the number of match victories in the tournament. Tournaments are series of head-to-head matches between two players, the winner of the round advancing to the next round. The first R rounds are best of three (i.e., win two matches to win the round, any unnecessary games are not played), the remaining rounds are best of five (i.e., win three matches to win the round, any unnecessary games are not played). Each tournament has a different value of R, but is not published.

Input Format

The first line is an integer N (1 <= N <= 100), the number of test cases, which follow. Each test case begins with a line containing an integer M (1 <= M <= 10). The following 2^M lines are of format "player_handle number_of_match_victories". Player handles are alphanumeric and between 1 and 16 characters long.

You may assume that the data provided describes a valid tournament.

Output Format

Print the player handles sorted in descending order of which round they survived to. For players who survived the same number of rounds, sort them lexicographically by player handle.

Sample Input

1
2
John 1
Jake 5
Joe 4
Jane 0

Sample Output

Jake
Joe
Jane
John

　　题意：给你2^m个选手的比赛胜利状况，这2^m个选手进行挑战赛，问最终按照进行比赛的round数来排序输出，若是round数相同的就按照字典序输出。其中一个round有可能三局两胜有可能五局三胜。(前R场三局两胜，剩下的五局三胜)
　　说实话第一次读题意的时候彻底看不懂，看了好几回才看懂，比赛的时候以为应该枚举R,由于m最大只有10，换而言之挑战赛构成的树最深只有10层，因此直接枚举R，若是有合法的状态就输出。结果不够时间敲，赛后听题解说的是R实际上是肯定的不须要枚举。
　　刚才用本身的想法实现了一下，WA了主要仍是没有肯定R。其实分析一下能够发现，对于全部的人，明显胜利场数最多的人一定是冠军(树根)他的round数绝对是最多的,而后就是胜利场数稍微少一点的绝对是比冠军少一round，而后继续这样推下去就能够发现若是咱们先按照胜利场数推下去的话咱们能够发现，场数多的绝对是round数多，每一次咱们能够肯定2^i我的排在前面，最后就能将2^m的人都排好。

上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <cstring>
 5 #include <utility>
 6 #include <vector>
 7 #include <queue>
 8 #include <algorithm>
 9 #define MAX 102
10 using namespace std;
11 
12 typedef pair<int,string> pii;
13 pii p;
14 vector<pii> u;
15 int m;
16 
17 bool cmp0(pii x,pii y){
18     return x.first==y.first ? x.second<y.second : x.first>y.first;
19 }
20 
21 bool cmp1(pii x,pii y){
22     return x.second<y.second;
23 }
24 
25 int main()
26 {
27     int t,n;
28     //freopen("data.txt","r",stdin);
29     ios::sync_with_stdio(false);
30     cin>>t;
31     while(t--){
32       cin>>m;
33       n=1<<m;
34       u.clear();
35       for(int i=0;i<n;i++){
36         cin>>p.second>>p.first;
37         u.push_back(p);
38       }
39       sort(u.begin(),u.end(),cmp0);
40       int bound=1;
41       for(vector<pii>::iterator it=u.begin()+1;it!=u.end();bound<<=1){
42             sort(it,it+bound,cmp1);
43             it=it+bound;
44       }
45       for(vector<pii>::iterator it=u.begin();it!=u.end();it++){
46           cout<<(*it).second<<endl;
47       }
48     }
49     return 0;
50 }

/*12450*/