题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1137
题目:
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.
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, for n subjects.
For each given data set, the program should write to standard output a line containing the result.
An example is given in Figure 1.
Input
7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0
Output
5
2
题目描述:
最大独立集,先二分匹配(匈牙利算法)求出匹配数,然后再用顶点数减去匹配数/2即可。代码:
#include<stdio.h>
#include<string.h>
int e[1010][1010],bo[1010],ma[1010];
int sum;
int a,b,c;
int dfs(int x)
{
int i;
for(i = 0;i < a;i ++){
if(!bo[i] && e[x][i]){
bo[i] = 1;
if(!ma[i] || dfs(ma[i])){
ma[i] = x;
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,w;
while(scanf("%d",&a) != EOF){
memset(e,0,sizeof(e));
for(i = 0;i < a;i ++){
scanf("%d: (%d)", &w, &b);
for(j = 0;j < b;j++){
scanf("%d",&c);
e[w][c] = 1;
}
}
sum = 0;
memset(ma,0,sizeof(ma));
for(j = 0;j < a;j ++){
memset(bo,0,sizeof(bo));
if(dfs(j))
sum ++;
}
printf("%d\n",a - sum/2);
}
return 0;
}