poj1703_Find them, Catch them_并查集
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 42451 | Accepted: 13059 |
Description
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
Output
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
#include <iostream>
#include <cstdio> #define MAX_N 1000000+5 using namespace std; int par[MAX_N];//父节点
int depth[MAX_N];//深度 void init(int n){
for(int i=;i<=n;i++){
par[i]=i;
depth[i]=;
}
}
int find_father(int t){
if(t==par[t]){
return t;
}else{
return par[t]=find_father(par[t]);
//实现了路径压缩
}
}
void unite(int t1,int t2){
int f1=find_father(t1);
int f2=find_father(t2);
if(f1==f2){
return ;
}
if(depth[f1]<depth[f2]){
par[f1]=f2;
}else{
par[f2]=f1;
if(depth[f1]==depth[f2]){
depth[f1]++;
//记录深度
}
}
} bool same(int x,int y){
return find_father(x)==find_father(y);
} int main()
{
int t;
int n,m;
char c;
int a,b;
scanf("%d",&t);
while(t--){
scanf("%d %d",&n,&m);
init(n*);
while(m--){
getchar();
scanf("%c %d %d",&c,&a,&b);
if(c=='D'){
unite(a,b+n);
unite(a+n,b);
}else{
if(same(a,b+n)){
printf("In different gangs.\n");
continue;
}
if(same(a,b)){
printf("In the same gang.\n");
continue;
}else{
printf("Not sure yet.\n");
continue;
}
}
}
}
return ;
}
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