并查集算法实战-优快云博客

Find them, Catch them

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

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.
几天不做题，都忘了并查集了。。。。
刚开始按数学思维胡搞的，想当然的WA了。。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int map[100001];
int main()
{
	int t;
	scanf("%d",&t);
	int a,b,n,m;
	char c[2];
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)
		map[i]=0;
		for(int i=1;i<=m;i++)
		{
			scanf("%s%d%d",&c,&a,&b);
			if(c[0]=='D')
			{
				if(map[a]==0&&map[b]==0)
					map[a]=1,map[b]=2;
				else if(map[a]==0&&map[b]!=0)
				map[a]=3-map[b];
				else
				map[b]=3-map[a];
					
			}
			else
			{
				if(map[a]==0||map[b]==0)
				printf("Not sure yet.\n");
				else if(map[a]==map[b])
				printf("In the same gang.\n");
				else
				printf("In different gangs.\n");
			}
		}
	}
	return 0;
 } 

并查集解法：
这题只有两个种类，也就是只有0和1两种， 对于两个不同的种类，那么之间的权值是相差1的，所以按照带权并查集的方法做加上1，然后取余2即可。
“*”符:用以表示该输入项,读入后不赋予相应的变量，即跳过该输入值。 
 如: 
 scanf("%d %*d %d",&a,&b); 
 当输入为：1 2 3时，把1赋予a，2被跳过，3赋予b。

     
      #include "stdio.h"
     
     
       
     
     
      int 
      main()
     
     
      {
     
     
          
      int 
      a, b;
     
     
       
     
     
          
      scanf
      (
      "%d%*c%d"
      , &a, &b);
     
     
       
     
     
          
      printf
      (
      "%d %d\n"
      , a, b);
     
     
       
     
     
          
      return 
      0;
     
     
      }
     
     
       
     
     
       
     
     
      你输入1n2，那么
      scanf
      把1读取赋给变量a，如果跳过一个
      char
      （这里是 ‘n’）接着读取2赋给变量b
     
     
      
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       #include<cstdio>
int f[100005],rank[100005];
int find(int x)
{
	if(x==f[x])
	return x;
	int t=f[x];
	f[x]=find(f[x]);
	rank[x]=(rank[x]+rank[t])%2;
	return f[x];
}
bool in(int x,int y)
{
	int a=find(x);
	int b=find(y);
	if(a==b)
	return false;
	f[b]=a;
	rank[b]=(rank[x]-rank[y]+1)%2;
}
int main()
{
int t,a,b;char c;
scanf("%d",&t);
while(t--)
{
	int n,m;
	scanf("%d%d%*c",&n,&m);
	for(int i=1;i<=n;i++)
	{
		f[i]=i;rank[i]=0;
	 } 
	 for(int i=1;i<=m;++i)
	 {
	 scanf("%c%d%d%*c",&c,&a,&b);
	 if(c=='D')
	 	in(a,b);
	else	
	{
	int aa= find(a), bb=find(b);  
                if(aa==bb)
		{  
                    if(rank[a]==rank[b])
                               printf("In the same gang.\n");  
                    else printf("In different gangs.\n");  
                }  
                else  
                    printf("Not sure yet.\n");  	
	  }
	 }
	}
	return 0;	
}