POJ 1703 带权并查集

Find them, Catch them

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47160		Accepted: 14521

DescriptionThe 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, giventwo 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. InputThe 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.OutputFor 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

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.

题意： n个人 m个关系，2个犯罪团伙，A x y :询问x和y的关系   D x y :x和y处于不同的团伙，

思路： 开一个rank数组记录该节点与根节点的关系  0：表示是一个团伙  1: 表示不是一个团伙。

当x和y的根节点不一样时，则表明他们的关系不能确定。

#include<stdio.h>#define N 100005using namespace std;int n,m;int f,rank;int getf(int x){if(f[x]==x) return x;int fa=getf(f[x]);rank[x]=(rank[x]+rank[f[x]])%2;f[x]=fa;return fa;}void merge(int x,int y){int t1=getf(x);int t2=getf(y);if(t1!=t2){f[t2]=t1;rank[t2]=(rank[x]-rank[y]+1+2)%2;}return ;}int jud(int x,int y){if((rank[x]+rank[y])%2) return 1;else return 0;}int main(){int i,j;int x,y;int ccase;scanf("%d",&ccase);while(ccase--){scanf("%d %d",&n,&m);for(i=0;i<=n;i++){f[i]=i;rank[i]=0;}char s[5];for(i=1;i<=m;i++){scanf("%s %d %d",s,&x,&y);if(s[0]=='D'){merge(x,y);}else{if(getf(x)!=getf(y)){printf("Not sure yet.\n");}else{if(jud(x,y)) printf("In different gangs.\n");else printf("In the same gang.\n");}}}}return 0;}