2018_2_12_Ubiquitous Religions_并查集
2018-02-12 14:03
260 查看
Ubiquitous Religions
DescriptionThere are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.
You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion. InputThe input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0. OutputFor each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.Sample Input10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
Sample OutputCase 1: 1
Case 2: 7
HintHuge input, scanf is recommended.SourceAlberta Collegiate Programming Contest 2003.10.18#include<iostream>
#include<cstdio>
#include<cstring>
#include<set>
#include<algorithm>
using namespace std;
const int N=5e5+10;
int sset
;
int set_find(int x){
if(sset[x]==-1)return x;
return sset[x]=set_find(sset[x]);
}
int main(){
int n,m;
int z=1;
while(scanf("%d%d",&n,&m),n||m){
memset(sset,-1,sizeof(sset));
int x,y;
while(m--){
scanf("%d%d",&x,&y);
x=set_find(x);
y=set_find(y);
if(x!=y){
sset[x]=y;
}
}
set<int> s;
for(int i=1;i<=n;i++)
s.insert(set_find(i));
printf("Case %d: %d\n",z++,s.size());
}
}
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 38587 | Accepted: 18391 |
You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion. InputThe input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0. OutputFor each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.Sample Input10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
Sample OutputCase 1: 1
Case 2: 7
HintHuge input, scanf is recommended.SourceAlberta Collegiate Programming Contest 2003.10.18#include<iostream>
#include<cstdio>
#include<cstring>
#include<set>
#include<algorithm>
using namespace std;
const int N=5e5+10;
int sset
;
int set_find(int x){
if(sset[x]==-1)return x;
return sset[x]=set_find(sset[x]);
}
int main(){
int n,m;
int z=1;
while(scanf("%d%d",&n,&m),n||m){
memset(sset,-1,sizeof(sset));
int x,y;
while(m--){
scanf("%d%d",&x,&y);
x=set_find(x);
y=set_find(y);
if(x!=y){
sset[x]=y;
}
}
set<int> s;
for(int i=1;i<=n;i++)
s.insert(set_find(i));
printf("Case %d: %d\n",z++,s.size());
}
}
相关文章推荐
- 2018_2_12_Building Bridges_world_final_并查集二维_缩点
- 2018_2_12_Network Connections_并查集
- 2018_2_12_Wireless Network_并查集
- 2018_2_12_War _并查集
- 2018_2_12_并查集二维
- 蓝桥杯 历届试题 PREV-12 危险系数 并查集找割点 Java
- 按键驱动-2018/01/12
- 2018_2_10_Find them, Catch them_并查集
- [NWPU][2014][TRN][12]并查集D - A Bug's Life POJ 2492
- COCI 2017/2018 Round #3,November 25th,2017 Pictionary [带权并查集+启发式合并]
- 安徽科技学院2017-2018-1学期15电信12班《Java编程技术》课下作业~解题报告
- 安徽科技学院2017-2018-1学期15电信12《Java编程技术》期末考试
- 2018_2_10_Cube Stacking_并查集进阶
- PKUACM 2018 D chocolate【并查集+克鲁斯卡尔】
- kettle 2018-01-12
- 每日一题(12)——计算Slim Span(并查集)
- 2017-2018-12 20155307 《信息安全系统设计基础》第十一周学习总结
- COCI 2017/2018 Round #2,November 4th,2017 K Usmjeri [LCA+并查集]
- 2018版OCP 11g 052最新考试题库整理(带答案)(12)
- 2018_2_14_This Sentence is False_并查集_关系