首页 > 代码库 > HDU4751Divide Groups【判断二分图】
HDU4751Divide Groups【判断二分图】
Divide Groups
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1141 Accepted Submission(s): 414
Problem Description
This year is the 60th anniversary of NJUST, and to make the celebration more colorful, Tom200 is going to invite distinguished alumnus back to visit and take photos.
After carefully planning, Tom200 announced his activity plan, one that contains two characters:
1. Whether the effect of the event are good or bad has nothing to do with the number of people join in.
2. The more people joining in one activity know each other, the more interesting the activity will be. Therefore, the best state is that, everyone knows each other.
The event appeals to a great number of alumnus, and Tom200 finds that they may not know each other or may just unilaterally recognize others. To improve the activities effects, Tom200 has to divide all those who signed up into groups to take part in the activity at different time. As we know, one‘s energy is limited, and Tom200 can hold activity twice. Tom200 already knows the relationship of each two person, but he cannot divide them because the number is too large.
Now Tom200 turns to you for help. Given the information, can you tell if it is possible to complete the dividing mission to make the two activity in best state.
Input
The input contains several test cases, terminated by EOF.
Each case starts with a positive integer n (2<=n<=100), which means the number of people joining in the event.
N lines follow. The i-th line contains some integers which are the id
of students that the i-th student knows, terminated by 0. And the id starts from 1.
Each case starts with a positive integer n (2<=n<=100), which means the number of people joining in the event.
N lines follow. The i-th line contains some integers which are the id
of students that the i-th student knows, terminated by 0. And the id starts from 1.
Output
If divided successfully, please output "YES" in a line, else output "NO".
Sample Input
33 01 01 2 0
Sample Output
YES
Source
2013 ACM/ICPC Asia Regional Nanjing Online
大意：有n个人，现在告诉你每个人认识哪些人，问你能不能把这些人分成两个集合，使每个集合中的人之间都相互认识（认识不可传递）
思路：用二分图来做：（晚上再写）
代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <vector> 5 using namespace std; 6 7 const int maxn = 105; 8 int G[maxn][maxn]; 9 vector<int> mat[maxn];10 int color[maxn];11 12 bool is_bi(int u) {13 for(int i = 0; i < mat[u].size(); i++) {14 int v = mat[u][i];15 if(color[v] == color[u]) return false;16 if(color[v] == 0) {17 color[v] = 3 - color[u];18 if(!is_bi(v)) return false;19 }20 }21 return true;22 }23 24 bool solve(int n) {25 for(int i = 1; i <= n; i++) {26 memset(color, 0, sizeof(color));27 color[i] = 1;28 if(!is_bi(i)) return false;29 }30 return true;31 }32 33 int main() {34 int n;35 int id;36 while(EOF != scanf("%d",&n)) {37 memset(G, 0, sizeof(G));38 memset(mat, 0, sizeof(mat));39 for(int i = 1; i <= n; i++) {40 mat[i].clear();41 while(scanf("%d",&id) && id) {42 G[i][id] = 1; 43 }44 }45 for(int i = 1; i <= n; i++) {46 for(int j = 1; j <= n; j++) {47 if(i == j) continue;48 if(G[i][j] + G[j][i] != 2) {49 // printf("%d %d\n",i, j);50 mat[i].push_back(j);51 mat[j].push_back(i);52 }53 }54 }55 if(solve(n)) puts("YES");56 else puts("NO");57 }58 return 0;59 }
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。