二分图最大独立集
来源:互联网 发布:黑色星期五抢购软件 编辑:程序博客网 时间:2024/05/01 21:29
二分图最大独立集=点数n-二分图最大匹配。知道这个公式后,这道题就很easy了,,是一道裸题,最基本的二分图最大独立集,,图都是直接建好得。。。题目:
Girls and Boys
Time Limit: 5000MS Memory Limit: 10000KTotal Submissions: 7452 Accepted: 3242
Description
In the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been "romantically involved". The result of the program is the number of students in such a set.
Input
The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:
the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)
The student_identifier is an integer number between 0 and n-1 (n <=500 ), for n subjects.
the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)
The student_identifier is an integer number between 0 and n-1 (n <=500 ), for n subjects.
Output
For each given data set, the program should write to standard output a line containing the result.
Sample Input
70: (3) 4 5 61: (2) 4 62: (0)3: (0)4: (2) 0 15: (1) 06: (2) 0 130: (2) 1 21: (1) 02: (1) 0
Sample Output
52ac代码:
#include <iostream>#include <cstdio>#include <algorithm>#include <string.h>#include <vector>using namespace std;vector<int> ss[505];int flag[505],visted[505];bool dfs(int x){for(int i=0;i<ss[x].size();++i){if(!visted[ss[x][i]]){ visted[ss[x][i]]=1; if(flag[ss[x][i]]==-1||dfs(flag[ss[x][i]])){ flag[ss[x][i]]=x;return true; }}}return false;}int main(){ int n; while(~scanf("%d",&n)){ memset(ss,0,sizeof(ss));memset(flag,-1,sizeof(flag));int x,num,y,m=n;char ch1,ch2,ch3;while(m--){ cin>>x>>ch1>>ch2>>num>>ch3; while(num--){ scanf("%d",&y); ss[x].push_back(y); }}int sum=0;for(int i=0;i<n;++i){ memset(visted,0,sizeof(visted)); if(dfs(i)) sum++;}printf("%d\n",n-sum/2); } return 0;}
- 二分图最大独立集
- 二分图最大独立集
- 二分图最大独立集
- 二分图最大匹配 & 最大独立集
- hdu 3829 二分图最大独立集
- poj 2771 二分图最大独立集
- HDU2768二分图求最大独立集
- poj3692 Kindergarten 二分图最大独立集
- hdu4160 Dolls (二分图最大独立集)
- poj 1466 二分图 最大独立集
- 二分图最大独立集--poj2771
- 二分图最大独立集-poj1466
- poj1466 二分图最大独立集
- HDU1054Strategic Game【二分图最大独立集】
- poj3020 二分图匹配 最大独立集
- 二分图(最大独立集)
- poj2771 二分图的最大独立集
- hdu3829 二分图的最大独立集
- 【转】无线连接两台笔记本有四种方法
- 各种数据库连接
- 【转】在网页中嵌入QQ
- 无聊 乱写写 CSDN博客挺好玩的 确实
- CSS面试题
- 二分图最大独立集
- 第一次负责项目总结
- 如何加快建 index 索引 的时间
- 【转】MSN,QQ在线即时交谈网页代码
- JSF学习总结篇
- 【转】在网页嵌入QQ+MSN+淘宝旺旺+Gtalk快速对话框官方代码的方法
- 输入自动完成
- 【转】在网页中嵌入QQ 阿里旺旺 淘宝代码及详解
- JS字符串累加