Find them, Catch them
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 49867 Accepted: 15320
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:
D [a] [b]
where [a] and [b] are the numbers of two criminals, and they belong to different gangs.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.
解题心得：
- 有两个帮派，龙帮和蛇帮，有n个罪犯，每个罪犯属于其中一个帮派，有m个操作，第一种就是规定罪犯a，b不属于一个帮派，第二种就是询问罪犯a，b的关系。
- 刚开始看这个题的时候真的很蒙，写得暴力会TLE，后来看大佬说有两种方法，
- 第一种，我把它称为创造镜像（a的镜像是a+n），每个罪犯的镜像和自己属于敌对关系，这样就可以把并查集分成两个部分，两个敌对关系的集合，如果给出a和b是敌对关系，那么就可以合并a和b镜像，合并b和a的镜像。
- 第二种是直接观察关系的转变，在find father和merge时候维护关系，首先使用cm[i]来表示第i个罪犯的帮派（值为0，1），先全部初始化为0，如果a和b是敌对关系，那么cm[b] = (cm[a]+1+cm[b])%2，在find father的时候需要从祖先节点开始从上往下维护，可以发现cm[son] = (cm[son] + cm[father])%2，这里可以仔细想想这个关系，要明白的是我们将所有的帮派初始化都为0。
第一种方法：
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn = 3e5+100;
int father[maxn],n,m,t;
int find(int x){
if(father[x] == x)
return x;
return father[x] = find(father[x]);
}
void merge(int x,int y) {
int fx = find(x);
int fy = find(y);
father[fy] = fx;
}
bool _same(int x,int y) {
return find(x) == find(y);
}
void unite(int x,int y) {
int fx = find(x);
int fy = find(y);
father[fy] = fx;
}
void init() {
scanf("%d%d",&n,&m);
for(int i=1;i<=n*3;i++)
father[i] = i;
}
void Solve() {
for(int i=0;i<m;i++) {
char temp[5];
int a,b;
scanf("%s%d%d",temp,&a,&b);
if(temp[0] == 'A') {
if(find(a) == find(b)) {
printf("In the same gang.
");
} else if(_same(a,b+n) && _same(a+n,b)) {
printf("In different gangs.
");
} else
printf("Not sure yet.
");
} else {
unite(a,b+n);
unite(a+n,b);
}
}
}
int main() {
scanf("%d",&t);
while(t--) {
init();
Solve();
}
return 0;
}第二种方法：
#include <algorithm>
#include <stdio.h>
#include <cstring>
using namespace std;
const int maxn = 1e5+100;
int father[maxn],n,m,cm[maxn];
void init() {
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
father[i] = i;
memset(cm,0,sizeof(cm));
}
int find(int x) {
if(x == father[x])
return x;
int temp = father[x];
father[x] = find(father[x]);
cm[x] = (cm[temp]+cm[x])%2;
return father[x];
}
void Unite(int a,int b) {
int fa = find(a);
int fb = find(b);
if(fa == fb)
return;
father[fb] = fa;
cm[fb] = (cm[a] + 1 - cm[b]) % 2;
}
int main() {
int t;
scanf("%d",&t);
while(t--){
init();
while(m--) {
char temp[5];
int a,b;
scanf("%s%d%d",temp,&a,&b);
if(temp[0] == 'A') {
if (find(a) != find(b)) {
printf("Not sure yet.
");
} else if (cm[a] == cm[b]) {
printf("In the same gang.
");
} else
printf("In different gangs.
");
} else {
Unite(a,b);
}
}
}
return 0;
}