POJ P1703 Find them, Catch them 并查集

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 26667 Accepted: 8084

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：有两个不同的帮派，龙帮和蛇帮。每个帮派至少有一个人。 判断两个人是否属于同一个帮派。有 T 组测试数据。 给你 N 个人，编号从 1 到 N，操作 M 次。每次操作输入一个字符和两个数 x ,y 如果字符为 A 则判断 x 和 y 是否属于同一个帮派，并且输出结果。如果字符为 D 则明确告诉你 x 和 y 是属于不同帮派的。

和上道题的思路完全一样。。。cin加优化一样会超时，所以用scanf输入数据。

#include <cstdio>int set[100010],offset[100010];int k,i,n,m,x,y,fx,fy;char s;int FindSet(int x);void Union(int x,int y,int fx,int fy);int main(){scanf("%d",&k);while (k--){scanf("%d %d",&n,&m);for (i=1;i<=n;++i){set[i]=i;offset[i]=0; }for (i=1;i<=m;++i){getchar();scanf("%c %d %d",&s,&x,&y);fx=FindSet(x);fy=FindSet(y);if (s == 'A')if (fx != fy)printf("Not sure yet.\n");elseif (offset[x] == offset[y])printf("In the same gang.\n");elseprintf("In different gangs.\n");elseUnion(x,y,fx,fy);}}return 0;}int FindSet(int x){if (set[x] != x){int temp=set[x];set[x]=FindSet(set[x]);offset[x]^=offset[temp];}return set[x];}void Union(int x,int y,int fx,int fy){set[fx]=fy;offset[fx]=offset[x]^offset[y]^1; return;}

kdwycz的网站：  http://kdwycz.com/

kdwyz的刷题空间：http://blog.csdn.net/kdwycz