10317 Fans of Footbal Teams

时间限制:1000MS 内存限制:65535K
提交次数:0 通过次数:0

题型: 编程题语言: G++;GCC

Description

Two famous football teams, named AC Milan(AC米兰) and Inter Milan(国际米兰) will have a match in GuangZhou City, which is

exciting. So a lot of fans get together to watch the wonderful match. This trouble the local polices. In order to avoid a

chaos caused by fans of the two teams, the police office in GuangZhou City decides to arrange the seats of the gymnasium(体育馆)

during the match. All fans of AC Milan seat Noth, while all fans of Inter Milan seat South . However, the police first needs

to identify which team a fan support. The present question is, given two fans; do they support a same team? You must give

your judgment based on incomplete information. 

Assume N (N <= 10^5) fans are currently in GuangZhou City, numbered from 1 to N. You will be given M (M <= 10^5) messages

in sequence, which are in the following two kinds: 

1. D [a] [b]

where [a] and [b] are the numbers of two fans, and they support different teams. 

2. A [a] [b]

where [a] and [b] are the numbers of two fans. This requires you to decide whether a and b support a same team.

输入格式

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow.

Each test case begins with a line with two integers N and M, followed by M lines each containing one message as described above.

输出格式

For each message "A [a] [b]" in each case, your program should give the judgment based on the information got before.

The answers might be one of "Support the same team.", "Support different teams." and "Not sure yet."

输入样例

输出样例

Not sure yet.

Support different teams.

Support the same team.

作者

admin

　　题意很好理解，，，就是给出信息语句告诉你哪两个人是支持不同队的，，，然后还有询问语句询问两个人是否是支持同一队伍；我解这题的解法是略有点奇葩的，，，算是套用了poj 1182(并查集经典题目，没做的话可以了解下)这道题的解法。。。既然是只提供信息说某两个人是支持不同的队伍，那么用 f[i]=i 表示"第i个人支持的队伍是跟某个人支持的队伍相同的"这一事件；而 f[i+n]=i+n 表示 "事件f[i]=i 的对立事件"；注意！这里的数组f[i]=i的意思是"编号i的人支持某支队伍"事件(不用理会他实际支持的是什么队)，而不是指编号为i的人！如果数据给了D 2 5的话，那就把f[2]和f[5+n]合并,f[2+n]和f[5]合并；因为如果编号2和5的人支持不同的队伍，那么f[2]和 "f[5]的对立事件"-f[5+n]必定是同时存在的，而f[2]的对立事件f[2+n]是和f[5]同时存在的。

上面的话理解可能有点绕,,,可以直接看代码：

 #include <iostream>

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 #include <cmath>

 #include <cstdlib>

 #include <cctype>

 #include <queue>

 #include <stack>

 #include <map>

 #include <vector>

 #include <set>

 #include <utility>

 #define ll long long

 #define inf 0x3f3f3f3f

 using namespace std;

 int t,n,m,temp;

 int f[];//记录父亲结点

 int father(int u) //寻找父亲

 {

     if(f[u]==u)

         return u;

     return f[u]=father(f[u]);

 }

 void unite(int x,int y)//合并x和y两个集合

 {

     int tx=father(x),ty=father(y);

     if(tx!=ty)

         f[ty]=tx;

 }

 bool same(int x,int y)//判断x和y是否属于同一集合

 {

     return father(x)==father(y);

 }

 int main()

 {

     //freopen("input.txt","r",stdin);

     scanf("%d",&t);

     while(t--)

     {

         char str[];

         int a,b;

         scanf("%d%d",&n,&m);

         temp=*n;

         for(int i=;i<=temp;i++)

             f[i]=i;

         while(m--)

         {

             scanf("%s%d%d",str,&a,&b);

             if(str[]=='D')

             {   //将a和b+n合并,b和a+n合并

                 unite(a,b+n);

                 unite(a+n,b);

             }

             else

             {

                 if(same(a,b+n))

                     printf("Support different teams.\n");

                 else if(same(a,b))

                     printf("Support the same team.\n");

                 else

                     printf("Not sure yet.\n");

             }

         }

     }

     return ;

 }

　　其实这题也可以不用这么傻逼的做法的,,,同宿舍的大神告诉我其实用结构体存数据，然后里面再加个数据域用来存放第i个人的对立队伍的集合编号就可以了