POJ 1703 Find them, Catch them

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 25909 Accepted: 7832

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.

Source

POJ Monthly--2004.07.18

 

题意：

有两个不同帮派

A a b表示询问成员a和b的帮派关系

D a b表示a和b在不同的帮派中

输出询问结果

 

代码：

# include <stdio.h># include <stdlib.h># include <string.h># include <math.h># define Max 100010int father[Max],e[Max];  //e[i]存储成员i的敌人int find (int a) {if (a!=father[a])father[a]=find(father[a]);return father[a];  }int main (){int t,n,m,i,a,b;char cmd[4];scanf ("%d",&t);while (t--) {scanf ("%d%d",&n,&m);memset (e,-1,sizeof (e));for (i=1;i<=n;i++)father[i]=i;for (i=0;i<m;i++){scanf ("%s%d%d",cmd,&a,&b);if (cmd[0]=='A') {if (find(a)==find(b))printf ("In the same gang.\n");else if (e[b]!=-1 && find(a)==find(e[b]))printf ("In different gangs.\n");else printf ("Not sure yet.\n");}else {if (e[a]!=-1) //若a存在敌人father[find(e[a])]=father[find(b)]; //b与a的敌人在同一帮派if (e[b]!=-1) //若b存在敌人father[find(e[b])]=father[find(a)]; //a与b的敌人在同一帮派e[a]=b; //更新a的敌人e[b]=a; //更新b的敌人}}}return 0;}

思路：

并查集

敌人的敌人是朋友