POJ - 1703 Find them, Catch them （并查集2）

Find them, Catch them

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27836 Accepted: 8486

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

——————————————————————————————————————————————

方法1：

将10^5看成10000……Runtime Error 一次……

这次并查集的对象是事件，而不仅仅是单一元素。在同一集合里表示该事件同时发生或同时不发生

D(x, y)： //分别合并两种集合，即（x在势力1，y在势力2）或（x在势力2， y在势力1）

unite(x, y+N);unite(x+N, y);

A(x, y)：

same(x, y+N) || same(x+N, y)表示： x在势力1，y在势力2 || x在势力2， y在势力1 。即x，y在不同势力

same(x, y) || same(x+N, y+N)表示： x，y都在势力1 || x，y都在势力2。 即x，y在相同势力

same(x, y+N)表示在不同的势力
same(x, y)表示在相同的势力

其他情况为 Not Sure.

#include <iostream>#include <cstdio>using namespace std;#define MAX 100010int cas, N, M;char sans[3][30] = {"In different gangs.\n","In the same gang.\n", "Not sure yet.\n"};//并查集int pa[MAX*2], rk[MAX*2];void init(int n) {for(int i = 0; i < n; i++){pa[i] = i; rk[i] = 0;}}int find(int x) { return pa[x] == x ? x : pa[x] = find(pa[x]); }void unite(int x, int y){x = find(x), y = find(y);if(x == y)return ;if(rk[x] < rk[y])pa[x] = y;else{pa[y] = x;if(rk[x] == rk[y])rk[x]++;}}int same(int x, int y){ return find(x) == find(y); }int main(){//freopen("in.txt","r",stdin); scanf("%d", &cas);while(cas--){scanf("%d %d", &N, &M);init(2*N);for(int i = 0; i < M; i++){char s[10];int x, y;scanf("%s%d%d", s, &x, &y);x--, y--;if(*s == 'A'){int ans = 2;if(same(x, y+N) /*|| same(x+N, y)*/)ans = 0;else if(same(x, y) /*|| same(x+N, y+N)*/) ans = 1;printf(sans[ans]);}else{unite(x, y+N);unite(x+N, y);}}}}

————————————————————————————————————————————————————

方法2：

A查询：先找出他们的老大：1、在同一集合，势力相同。 2、x的老大是y老大的对立面，不同势力。3、未确定。

D操作：若尚无敌对势力则记录，然后各自合并入敌对势力。

#include <iostream>#include <cstdio>using namespace std;#define MAX 100010int cas, N, M;int opp[MAX];char sans[3][30] = {"In different gangs.\n","In the same gang.\n", "Not sure yet.\n"};//并查集int pa[MAX*2], rk[MAX*2];void init(int n) {for(int i = 0; i < n; i++){pa[i] = i; rk[i] = 0; opp[i] = -1;}}int find(int x) { if(x==-1)return -1; return pa[x] == x ? x : pa[x] = find(pa[x]); }void unite(int x, int y){x = find(x), y = find(y);if(x == y)return ;if(rk[x] < rk[y])pa[x] = y;else{pa[y] = x;if(rk[x] == rk[y])rk[x]++;}}int same(int x, int y){ return find(x) == find(y); }int main(){//freopen("in.txt","r",stdin); scanf("%d", &cas);while(cas--){scanf("%d %d", &N, &M);init(N);for(int i = 0; i < M; i++){char s[10];int x, y;scanf("%s%d%d", s, &x, &y);x--, y--;int x1 = find(x), y1 = find(y);//先找各势力的老大（父节点）if(*s == 'A'){int ans = 2;if(x1 == find(opp[y1]))ans = 0;else if(x1 == y1) ans = 1;printf(sans[ans]);}else{//若尚无敌对势力则记录。if(opp[x] == -1)opp[x] = y;if(opp[y] == -1)opp[y] = x;//与敌对势力合并unite(x, opp[y1]);unite(y, opp[x1]);}}}}