HDU 1068 Girls and Boys(二分图最大独立点集,3级)
2013-07-19 16:47
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Girls and Boys
Time Limit:10000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
Submit Status
Appoint description:
System Crawler (2012-10-30)
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.
Input
Output
思路:ans=n-二分图最大独立点集/2
Time Limit:10000MS Memory Limit:32768KB 64bit IO Format:%I64d
& %I64u
Submit Status
Appoint description:
System Crawler (2012-10-30)
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.
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
思路:ans=n-二分图最大独立点集/2
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; const int mm=1e4+9; const int mn=5e6+9; class node { public:int v,next; }e[mn]; int head[mm],edge,n,dot; int Left[mm]; bool vis[mm]; void data() { memset(head,-1,sizeof(head));dot=edge=0; } void add(int u,int v) { e[edge].v=v;e[edge].next=head[u];head[u]=edge++; } void read() { int u,v,m;char c; scanf("%d:",&u); if(u==-1)u=0; while(1) { c=getchar(); if(c=='(')break; } scanf("%d",&m);dot=max(dot,u); while(1) { c=getchar(); if(c==')')break; } for(int i=0;i<m;++i) { scanf("%d",&v);dot=max(dot,v); if(u<v){add(u,v);add(v,u);} } } int dfs(int u) { int v; for(int i=head[u];~i;i=e[i].next) { v=e[i].v; if(vis[v])continue; vis[v]=1; if(Left[v]==-1||dfs(Left[v])) { Left[v]=u;return 1; } } return 0; } int get_max() { int ret=0; memset(Left,-1,sizeof(Left)); for(int i=0;i<=dot;++i) { memset(vis,0,sizeof(vis)); if(dfs(i)) ++ret; } return ret; } int main() { int u,v,m;char c; while(~scanf("%d",&n)) { data(); for(int i=0;i<n;++i) { read(); } int ans=n-get_max()/2; printf("%d\n",ans); } return 0; }
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