ZOJ1137 二分图的最大独立点集 匈牙利算法

Girls and Boys

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Time Limit: 10 Seconds      MemoryLimit: 32768 KB

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the second year of the university somebody started a study on the romanticrelations between the students. The relation ��romantically involved�� is defined between one girl and one boy. For the studyreasons 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 insuch a set.

The input contains several data sets in text format. Each data set representsone 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_identifier1student_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 linecontaining 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

 

求二分图的最大独立点集,用匈牙利算法先求出最大匹配数,然后用总数减去最大匹配数就是题目要求的答案.刚开始用了邻接矩阵,严重超时,改成邻接表就过了.

 

代码如下:

 

#include<stdio.h>#include<string.h>#define MAXN 500#define inf 2100000000bool visx[MAXN],visy[MAXN];int g[MAXN][MAXN];int pre[MAXN];int n,res;struct node{ int nod; node *next;} head[MAXN];int find(int deal){ int i,t; node *p; p=head[deal].next; while (p!=NULL) { if (visy[p->nod]) {p=p->next;continue;} visy[p->nod]=1; if (pre[p->nod]==-1||find(pre[p->nod])) { pre[p->nod]=deal; return 1; } p=p->next; } return 0;}int main(){ int i,k,m,j,nod,count; char s[20]; node *p; while (scanf("%d",&n)!=EOF) { for (i=1; i<=n; ++i) head[i].next=NULL; for (i=1; i<=n; ++i) { memset(s,0,sizeof(s)); scanf("%s",s); sscanf(s,"%d/n",&k); scanf("%s",s); sscanf(s,"(%d)",&m); ++k; for (j=1; j<=m; ++j) { scanf("%d",&nod); ++nod; p=new node; p->nod=nod; p->next=head[k].next; head[k].next=p; } } count=0; memset(pre,-1,sizeof(pre)); for (i=1; i<=n; ++i) { memset(visy,0,sizeof(visy)); if (find(i)) count++; } printf("%d/n",n-count/2); } return 0;}