POJ 1703 Find them, Catch them

来源:互联网 发布:mac电脑flash过期 编辑:程序博客网 时间:2024/06/16 02:20

Find them, Catch them
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 47170 Accepted: 14526

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

15 5A 1 2D 1 2A 1 2D 2 4A 1 4

Sample Output

Not sure yet.In different gangs.In the same gang.
题目大意:有两个黑帮团伙,给你几个人和两个操作,D表示这两个人不是一个团伙的,A表示询问这两个人是不是一个团伙的

这是一道很基础的带权并查集,比起食物链,这算是水题了

#include<iostream>#include<cstdio>using namespace std;int tree[100010];int Class[100010];int find(int x){if(x==tree[x]) return x;int tmp=tree[x];tree[x]=find(tree[x]);Class[x]=(Class[x]+Class[tmp])%2;return tree[x];}void merge(int a,int b){int x=find(a);int y=find(b);if(x!=y) tree[x]=y;Class[x]=(Class[a]+Class[b]+1)%2;}int main(){//freopen("in.txt","r",stdin);int t;cin>>t;while(t--){int n,m;scanf("%d%d",&n,&m);getchar();for(int i=0;i<=n;i++){tree[i]=i;Class[i]=0;}while(m--){char ch[3];int a,b;scanf("%s%d%d",ch,&a,&b);getchar();if(ch[0]=='A'){if(find(a)==find(b)){if(Class[a]!=Class[b]) printf("In different gangs.\n");else printf("In the same gang.\n");}else printf("Not sure yet.\n");}else merge(a,b);}}}



原创粉丝点击