[并查集] POJ 1703 Find them, Catch them
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 43132 | Accepted: 13257 |
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<stdio.h>
int father[150500],relation[150050];
void init(int n)
{
int i;
for(i=1;i<=n;++i) father[i]=i;
for(i=1;i<=n;++i) relation[i]=0;
}
int find(int x)
{
int t;
if(x==father[x]) return (x);
t=father[x];
father[x]=find(t);
relation[x]=(relation[x]+relation[t])%2;
return (father[x]);
}
void merge(int x,int y)
{
int find_x=find(x);
int find_y=find(y);
if(find_x!=find_y)
{
relation[find_x]=(relation[y]+1-relation[x]+2)%2;
father[find_x]=find_y;
}
}
int main()
{
int T,n,m,num=0;char c;int x,y;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
init(n);
while(m--)
{
getchar();
scanf("%c%d%d",&c,&x,&y);
if(c=='D')
{
merge(x,y);
}else
if(c=='A')
{
if(find(x)!=find(y))
{
printf("Not sure yet.\n");
continue;
}else
if(relation[x]==relation[y])
{
printf("In the same gang.\n");
continue;
}else
if(relation[x]!=relation[y])
printf("In different gangs.\n");
}
}
}
return 0;
}
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