首页 > ACM题库 > HDU-杭电 > HDU 1068 Girls and Boys-分治-[解题报告] C++
2013
11-28

HDU 1068 Girls and Boys-分治-[解题报告] C++

Girls and Boys

问题描述 :

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.

样例输入:

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

样例输出:

5
2

找出没有缘分的同学,如果有缘分的建边,

就是求最大独立集问题了





#include<stdio.h>
#include<string.h>
int n,m,ma[1050][1050],mark[1050],link[1050];
int find(int a)
{
    int i;
    for(i=0;i<n;i++)
    {
        if(ma[a][i]==1&&mark[i]==0)
        {
            mark[i]=1;
            if(link[i]==-1||find(link[i])==1)
            {
                link[i]=a;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,x,j,y,z,sum,temp=0;
    char ch1,ch2,ch3,ch4;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        memset(ma,0,sizeof(ma));
        memset(link,-1,sizeof(link));
        for(i=1;i<=n;i++)
        {
            scanf("%d%c%c%c%d%c",&x,&ch1,&ch2,&ch3,&z,&ch4);
            for(j=1;j<=z;j++)
            {
                scanf("%d",&y);
                 ma[x][y]=1;
            }
        }
        for(i=0;i<n;i++)
            {
                    memset(mark,0,sizeof(mark));
                      sum=sum+find(i);
            }
        printf("%d\n",n-sum/2);
    }
    return 0;
}


  1. #include <cstdio>

    int main() {
    //answer must be odd
    int n, u, d;
    while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
    if(n<=u) { puts("1"); continue; }
    n-=u; u-=d; n+=u-1; n/=u;
    n<<=1, ++n;
    printf("%dn",n);
    }
    return 0;
    }

  2. 思路二可以用一个长度为k的队列来实现,入队后判断下队尾元素的next指针是否为空,若为空,则出队指针即为所求。