POJ 1703

最新推荐文章于 2021-02-26 21:06:24 发布

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Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46068		Accepted: 14176

Description

The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon and Gang Snake. However, the police first needs to identify which gang a criminal belongs to. The present question is, given two criminals; do they belong to a same clan? You must give your judgment based on incomplete information. (Since the gangsters are always acting secretly.)

Assume N (N <= 10^5) criminals are currently in Tadu City, numbered from 1 to N. And of course, at least one of them belongs to Gang Dragon, and the same for Gang Snake. 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 criminals, and they belong to different gangs.

2. A [a] [b]
where [a] and [b] are the numbers of two criminals. This requires you to decide whether a and b belong to a same gang.

Input

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.

Output

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 "In the same gang.", "In different gangs." and "Not sure yet."

Sample Input

Sample Output

Not sure yet.
In different gangs.
In the same gang.

解题思路：一道并查集的题目，但是有点不一样，就是那个D，描述的是不同的，想了好久不知道如何处理，看了下别人的解题报告
有些地方还是不理解，以后有时间再看看！

#include<stdio.h>

int father[150500],relation[150050];

void init(int n)
  {
     int i;
     for(i=1;i<=n;++i) father[i]=i;
     for(i=1;i<=n;++i) relation[i]=0;
  }
  
int find(int x)
  {
     int t;
     if(x==father[x]) return (x);
     t=father[x];
     father[x]=find(t);
     relation[x]=(relation[x]+relation[t])%2;
     return (father[x]);
  }
  
void merge(int x,int y)
  {
     int find_x=find(x);
     int find_y=find(y);
     if(find_x!=find_y)
       {
          relation[find_x]=(relation[y]+1-relation[x]+2)%2;
          father[find_x]=find_y;
       }
  }
  
int main()
  {
     int T,n,m,num=0;char c;int x,y;
     scanf("%d",&T);
     while(T--)
       {
          scanf("%d%d",&n,&m);
          init(n);
          while(m--)
            {
              getchar();
              scanf("%c%d%d",&c,&x,&y);
              if(c=='D')
                {
                   merge(x,y);
                }else
              if(c=='A')
                {
                  if(find(x)!=find(y))
                    {
                       printf("Not sure yet.\n");
                       continue;
                    }else
                  if(relation[x]==relation[y])
                    {
                       printf("In the same gang.\n");
                       continue;
                    }else
                  if(relation[x]!=relation[y])
                    printf("In different gangs.\n");
                }
            }
       }
     return 0;
  }