HDU——1068 Girls and Boys
阿新 • • 發佈:2017-08-24
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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 Output
5
2
Girls and Boys
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12246 Accepted Submission(s): 5768
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 ...
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 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
Source Southeastern Europe 2000 二分圖的最大匹配數(裸題)
#include<cstdio> #include<cstdlib> #include<cstring> #include<iostream> #include<algorithm> #define N 1010 using namespace std; bool vis[N]; int t,n,m,x,y,ans,girl[N],map[N][N]; int read() { int x=0,f=1; char ch=getchar(); while(ch<‘0‘||ch>‘9‘) {if(ch==‘-‘) f=-1; ch=getchar();} while(ch<=‘9‘&&ch>=‘0‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } int find(int x) { for(int i=1;i<=n;i++) { if(!vis[i]&&map[x][i]) { vis[i]=true; if(girl[i]==-1||find(girl[i])) {girl[i]=x;return 1;} } } return 0; } int main() { while(scanf("%d",&n)!=EOF) { ans=0; memset(map,0,sizeof(map)); memset(girl,-1,sizeof(girl)); for(int i=1;i<=n;i++) { x=read();m=read(); while(m--) y=read(),map[x+1][y+1]=1; } for(int i=1;i<=n;i++) { memset(vis,0,sizeof(vis)); ans+=find(i); } printf("%d\n",(2*n-ans)/2); } return 0; }
HDU——1068 Girls and Boys