Find them, Catch them
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 31102 | Accepted: 9583 |
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.
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
题目大意：
T组測试数据，n个人。m组询问，D a b 表示 a,b 不在同一个gang（尽管不知道gang是什么意思？） ，A a b表示a和b的关系。
解题思路：
仅仅须要并查集，再增加一个enemy数组记录某人的一个敌人就可以。
解题代码：
#include <iostream>
#include <cstdio>
using namespace std; const int maxn=110000;
int father[maxn],enemy[maxn],n,m; int find(int x){
if(father[x]!=x){
father[x]=find(father[x]);
}
return father[x];
} void combine(int x,int y){
int a=find(x),b=find(y);
father[b]=a;
} void solve(){
char ch;
int a,b;
while(m-- >0){
getchar();
scanf("%c%d%d",&ch,&a,&b);
//cout<<ch<<"->"<<a<<"->"<<b<<endl;
if(ch=='D'){
if(enemy[a]!=-1) combine(enemy[a],b);
if(enemy[b]!=-1) combine(enemy[b],a);
enemy[a]=b;
enemy[b]=a;
}else{
if(enemy[a]==-1 || enemy[b]==-1 ) printf("Not sure yet.\n");
else{
if(find(a)==find(b)) printf("In the same gang.\n");
else if(find(enemy[a])==find(b) || find(a)==find(enemy[b]) ) printf("In different gangs.\n");
else printf("Not sure yet.\n");
}
}
}
} int main(){
int t;
scanf("%d",&t);
while(t-- >0){
scanf("%d%d",&n,&m);
for(int i=0;i<=n;i++){
father[i]=i;
enemy[i]=-1;
}
solve();
}
return 0;
}