题目:

  Here is World Cup again, the top 32 teams come together to fight for the World Champion. The teams are assigned into 8 groups, with 4 teams in each group. Every two teams in the same group will play a game (so there are totally 6 games in each group), and the winner of this game gets 3 points, loser gets 0 point. If it is a tie game, both teams get 1 point. After all games finished, we get the scoreboard, but we forget the result of each game, can you help us to figure the result of each game? We only care about the win/lose/tie result of each game, but we don’t care the goals in each game.

Input:

  The input starts with one line containing exactly one integer T, which is the number of test cases. Each test case contains four space-separated integers A, B, C, D, in a line, which indicate the points each team gets after all 6 games.

Output:

  For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is “Yes” if you can point out the result of each game, or “No” if there are multiple game results satisfy the scoreboard, or “Wrong Scoreboard” if there is no game result matches the scoreboard.

Note:

  In sample case #1, the only scenaro will be: the first team wins all the three games it plays, the second team loses to the first team and wins the other two, the third team only wins the game with the fourth, and the fourth team lose all the games.

  In sample case #2, the fourth team loses all the games, and the first three teams get into a winning-cycle, but there may be two different winning-cycles: first team wins second team, second team wins third team, third team wins first team OR first team wins third team, third team wins second team, second team wins first team. We can’t figure which winning-cycle is the actual game result.

  In sample case #3, the first team get 10 points, but no team could get more than 9 points by play three games, so it is a wrong scoreboard.

题意:世界杯比赛的一个小组中有4支队伍,队伍两两之间要进行一场比赛,赢得队伍得3分,输的队伍得0分,平局双方得1分。给出四支队伍的比赛完后的得分情况,问能不能确定具体的输赢情况。

思路:每组比赛一共有如下六种情况,那就dfs枚举六种情况下每个队伍输赢的所有的分情况,如果和给出的得分情况有多组相同,就是“No”,一组相同就是“Yes”,没有相同情况就是“Wrong Scoreboard”。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 1e5+;
int a,b,c,d,ans;
int buf[]; void dfs(int x)
{
if(x == )
{
if(a == buf[] && b==buf[] && c==buf[]&& d==buf[])
ans++;
return;
}
if(x==)
{
buf[]+=;//1队伍赢
dfs(x+);
buf[]-=;
buf[]+=,buf[]+=;//两队平局
dfs(x+);
buf[]-=,buf[]-=;
buf[]+=;//2队伍赢
dfs(x+);
buf[]-=;
}
else if(x==)
{
buf[]+=;//1队伍赢
dfs(x+);
buf[]-=;
buf[]+=,buf[]+=;//两队平局
dfs(x+);
buf[]-=,buf[]-=;
buf[]+=;//3队伍赢
dfs(x+);
buf[]-=;
}
else if(x==)
{
buf[]+=;//1队伍赢
dfs(x+);
buf[]-=;
buf[]+=,buf[]+=;//两队平局
dfs(x+);
buf[]-=,buf[]-=;
buf[]+=;//4队伍赢
dfs(x+);
buf[]-=;
}
else if(x==)
{
buf[]+=;//2队伍赢
dfs(x+);
buf[]-=;
buf[]+=,buf[]+=;//两队平局
dfs(x+);
buf[]-=,buf[]-=;
buf[]+=;//3队伍赢
dfs(x+);
buf[]-=;
}
else if(x==)
{
buf[]+=;//2队伍赢
dfs(x+);
buf[]-=;
buf[]+=,buf[]+=;//两队平局
dfs(x+);
buf[]-=,buf[]-=;
buf[]+=;//4队伍赢
dfs(x+);
buf[]-=;
}
else if(x==)
{
buf[]+=;//3队伍赢
dfs(x+);
buf[]-=;
buf[]+=,buf[]+=;//两队平局
dfs(x+);
buf[]-=,buf[]-=;
buf[]+=;//4队伍赢
dfs(x+);
buf[]-=;
}
} int main()
{
int T,cnt = ;
scanf("%d",&T);
while(T--)
{
ans = ;
memset(buf,,sizeof(buf));
scanf("%d%d%d%d",&a,&b,&c,&d);
dfs();
if(ans==)
printf("Case #%d: Wrong Scoreboard\n",cnt++);
else if(ans == )
printf("Case #%d: Yes\n",cnt++);
else
printf("Case #%d: No\n",cnt++);
}
return ;
}

World Cup(The 2016 ACM-ICPC Asia China-Final Contest dfs搜索)的更多相关文章

  1. 2016 ACM/ICPC Asia Regional Dalian Online 1006 /HDU 5873

    Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)To ...

  2. HDU 5873 Football Games 【模拟】 (2016 ACM/ICPC Asia Regional Dalian Online)

    Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)To ...

  3. 2016 ACM/ICPC Asia Regional Shenyang Online 1003/HDU 5894 数学/组合数/逆元

    hannnnah_j’s Biological Test Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K ...

  4. 2016 ACM/ICPC Asia Regional Qingdao Online 1001/HDU5878 打表二分

    I Count Two Three Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  5. 2016 ACM/ICPC Asia Regional Shenyang Online 1009/HDU 5900 区间dp

    QSC and Master Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) ...

  6. 2016 ACM/ICPC Asia Regional Shenyang Online 1007/HDU 5898 数位dp

    odd-even number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)T ...

  7. 2016 ACM/ICPC Asia Regional Dalian Online 1002/HDU 5869

    Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K ( ...

  8. HDU 5874 Friends and Enemies 【构造】 (2016 ACM/ICPC Asia Regional Dalian Online)

    Friends and Enemies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Othe ...

  9. HDU 5889 Barricade 【BFS+最小割 网络流】(2016 ACM/ICPC Asia Regional Qingdao Online)

    Barricade Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total S ...

  10. HDU 5875 Function 【倍增】 (2016 ACM/ICPC Asia Regional Dalian Online)

    Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total ...

随机推荐

  1. luogu3605晋升者计数

    https://www.zybuluo.com/ysner/note/1282069 题面 给一颗带点权的树,求每个点的子树中比该点权值大的点的个数. \(n\leq10^5\) 解析 首先有个很无脑 ...

  2. ubuntu 12.04.5 LTS版本 更新 source.list

    更新后一定要:apt-get update # # deb cdrom:[Ubuntu-Server LTS _Precise Pangolin_ - Release amd64 (20140806. ...

  3. bzoj2060

    树形dp dp[x][0]表示x点父亲没选,dp[x][1]表示x点父亲选了,然后dp[x][0]=max(sigma(dp[c[x]][0]),sigma(dp[c[x]][1])) dp[x][1 ...

  4. bzoj1770: [Usaco2009 Nov]lights 燈(折半搜索)

    1770: [Usaco2009 Nov]lights 燈 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 1153  Solved: 564[Submi ...

  5. Java经典算法之折半查找(二分法)

    采用二分法时,数据应是有序并且不重复的 与小时候玩的猜数游戏是一样的,会让你猜一个他所想的1~100之间的数,当你猜了一个数后,他会告诉你三种选择中的一个,比他想的大,或小,或猜中了,为了能用最少的次 ...

  6. 个人微信号二次开发SDK协议,个人微信号二次开发api接口

    通过这个API接口可以做什么? 通过我们提供的API接口您可以开发: 工作手机(如:X创,X码,XX管家等) 微信群讲课软件(如:讲课X师,一起X堂等) 微信社群管理软件(如:小X管家,微X助手等) ...

  7. nginx 多进程 + io多路复用 实现高并发

    一.nginx 高并发原理 简单介绍:nginx 采用的是多进程(单线程) + io多路复用(epoll)模型 实现高并发 二.nginx 多进程 启动nginx 解析初始化配置文件后会 创建(for ...

  8. liunx 用户切换 su sudo

    liunx 用户操作#useradd test#passwd test 用户身份切换su 切换用户 需要知道切换用户的密码1.su [-lm] [-c命令] [username] #su -login ...

  9. 03—AOP基本配置

  10. js实现左右点击图片层叠滚动特效

    需要加载js有 <script type="text/javascript" src="js/jquery-1.7.2.min.js"></s ...