Find them, Catch them
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 21467 |
|
Accepted: 6371 |
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
1
5 5
A 1 2
D 1 2
A 1 2
D 2 4
A 1 4
Sample Output
Not sure yet.
In different gangs.
In the same gang.
Source
POJ Monthly--2004.07.18囧，知道是并查集
也知道是根據路徑長度判斷是不是一個集合
但是 剛開始發現不能路徑壓縮，然后就裸的了，就tle了
然后………
寫了一串不知道是什么東西的東西，然后就過了
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iostream>
#include <sstream>
#include <fstream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <algorithm>
#include <iomanip>
#define maxn 100005
using namespace std;
int father[maxn];
int mark[maxn];
int n,m;
int find(int x)
{
if(father[x]==x) return x;
else
{
int pa=father[x];
father[x]=find(father[x]);
mark[x]=!(mark[x]^mark[pa]);
return father[x];
}
}
void unit(int x,int y)
{
int r1,r2;
r1=find(x);
r2=find(y);
father[r1]=r2;
mark[r1]=!((!(mark[x]^0))^mark[y]);
}
void cas_init()
{
for(int i=1; i<=maxn; i++) father[i]=i,mark[i]=1;
}
int main()
{
int x,y,t,tmp1,tmp2,len1,len2;
char str[5];
scanf("%d",&t);
while(t--)
{
cas_init();
scanf("%d%d",&n,&m);
for(int i=1; i<=m; i++)
{
scanf("%s%d%d",str,&x,&y);
if(str[0]=='A')
{
tmp1=find(x);
tmp2=find(y);
if(tmp1==tmp2)
{
if(mark[x]==mark[y])
printf("In the same gang.\n");
else printf("In different gangs.\n");
}
else printf("Not sure yet.\n");
}
else if(str[0]=='D')
{
unit(x,y);
}
}
}
return 0;
}