hdu 1068 Girls and Boys(最大独立集·maxmatch匈牙利)

题目：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19391

Girls and Boys

Time Limit: 10000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

Submit Status

Description

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. 

 

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

52 

 

Source

Southeastern Europe 2000

Submit Status

求不属于匹配关系的人数。
二分最大匹配后的匹配数是结果的2倍：

最大独立集=总个数-最大匹配数（U=n-maxmatch+maxmatch/2=n-maxmatch/2，n-maxmatch是绝对的未匹配，+maxmatch/2是加上完全拆开后的独立个数）。综上，敲代码。。。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=1e3+5;bool bmap[maxn][maxn],bmask[maxn];int nx,ny;int cx[maxn],cy[maxn];int times=0;int findpath(int u){    for(int j=1;j<=ny;j++){        if(bmap[u][j] && !bmask[j]){             bmask[j]=1;             if(cy[j]==-1 || findpath(cy[j])){                  cy[j]=u;                  cx[u]=j;                  return 1;             }        }    }    return 0;}int maxmatch(){    int res=0;    memset(cx,-1,sizeof(cx));    memset(cy,-1,sizeof(cy));    for(int i=1;i<=nx;i++){        if(cx[i]==-1){             for(int j=1;j<=ny;j++)   bmask[j]=0;             res+=findpath(i);        }    }    return res;}int main(){    //freopen("cin.txt","r",stdin);    int n;    while(cin>>n){        nx=ny=n;        memset(bmap,0,sizeof(bmap));        int ix,iy,rela;        while(n--){             scanf("%d",&ix);             while(getchar()!='(');             scanf("%d)",&rela);             while(rela--){                 scanf("%d",&iy);                 bmap[ix+1][iy+1]=1;             }        }        int ans=maxmatch();        ans=nx-ans+ans/2;        printf("%d\n",ans);    }    return 0;}