POJ 1703 Find them, Catch them (数据结构-并查集)
2014-07-30 16:18
495 查看
Find them, Catch them
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
Sample Output
Source
POJ Monthly--2004.07.18
题目大意:
T组测试数据,n个人,m组询问,D a b 表示 a,b 不在同一个gang(虽然不知道gang是什么意思?) ,A a b表示a和b的关系。
解题思路:
只需要并查集,再加入一个enemy数组记录某人的一个敌人即可。
解题代码:
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 31102 | Accepted: 9583 |
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
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.
Source
POJ Monthly--2004.07.18
题目大意:
T组测试数据,n个人,m组询问,D a b 表示 a,b 不在同一个gang(虽然不知道gang是什么意思?) ,A a b表示a和b的关系。
解题思路:
只需要并查集,再加入一个enemy数组记录某人的一个敌人即可。
解题代码:
#include <iostream> #include <cstdio> using namespace std; const int maxn=110000; int father[maxn],enemy[maxn],n,m; int find(int x){ if(father[x]!=x){ father[x]=find(father[x]); } return father[x]; } void combine(int x,int y){ int a=find(x),b=find(y); father[b]=a; } void solve(){ char ch; int a,b; while(m-- >0){ getchar(); scanf("%c%d%d",&ch,&a,&b); //cout<<ch<<"->"<<a<<"->"<<b<<endl; if(ch=='D'){ if(enemy[a]!=-1) combine(enemy[a],b); if(enemy[b]!=-1) combine(enemy[b],a); enemy[a]=b; enemy[b]=a; }else{ if(enemy[a]==-1 || enemy[b]==-1 ) printf("Not sure yet.\n"); else{ if(find(a)==find(b)) printf("In the same gang.\n"); else if(find(enemy[a])==find(b) || find(a)==find(enemy[b]) ) printf("In different gangs.\n"); else printf("Not sure yet.\n"); } } } } int main(){ int t; scanf("%d",&t); while(t-- >0){ scanf("%d%d",&n,&m); for(int i=0;i<=n;i++){ father[i]=i; enemy[i]=-1; } solve(); } return 0; }
相关文章推荐
- poj 1703 Find them, Catch them(数据结构:并查集)
- POJ 1703 Find them, Catch them [数据结构-并查集 union-find sets]
- POJ 1703 Find them, Catch them(数据结构-并查集)
- POJ 1703 Find them, Catch them (数据结构-并查集)
- POJ 1703 Find them, Catch them (数据结构-并查集)
- POJ 1703 Find them, Catch them 种类并查集
- POJ 1703 Find them, Catch them【典型并查集:判断在不同的集合】
- POJ 1703 Find them, Catch them(并查集,等价关系)
- poj 1703 Find them, Catch them (分组并查集 偏移向量实现)
- POJ 1703 Find them, Catch them(种类并查集)
- POJ 1703 Find them, Catch them(种类并查集)
- POJ 1703 Find them, Catch them (并查集)
- Poj 1703 Find them, Catch them【关系并查集】
- POJ 1703 Find them, Catch them(并查集)
- POJ 1703 Find them, Catch them【带权并查集】
- (用树结构支持并查集8.2.2)POJ 1703 Find them, Catch them(并查集的简单使用: 判断两个元素是否属于同一集合)
- POJ1703--Find them, Catch them--并查集
- POJ 1703 Find them, Catch them 种类并查集
- POJ:1703 Find them, Catch them(种类并查集(影子并查集))
- POJ 1703 Find them, Catch them【并查集】