UOFTCG - Office Mates
no tags
Dr. Baws has an interesting problem. His N
graduate students, while friendly with some select people, are
generally not friendly with each other. No graduate student is willing
to sit beside a person they aren‘t friends with.
The desks are up against the wall, in a single line, so it‘s possible
that Dr. Baws will have to leave some desks empty. He does know which
students are friends, and fortunately the list is not so long: it turns
out that for any subset of K
graduate students, there are at most K?1
pairs of friends. Dr. Baws would like you to minimize the total number of desks required. What is this minimum number?
Input
The input begins with an integer T≤50
, the number of test cases. Each test case begins with two integers on their own line: N≤100000, the number of graduate students (who are indexed by the integers 1 through N), and M, the number of friendships among the students. Following this are M lines, each containing two integers i and j separated by a single space. Two integers i and j represent a mutual friendship between students i and j
.
The total size of the input file does not exceed 2 MB.
Output
For each test case output a single number: the minimum number of desks Dr. Baws requires to seat the students.
Example
Input:
16 51 21 31 44 54 6
Output:
7
Explanation of Sample:
As seen in the diagram, you seat the students in two groups of three with one empty desk in the middle.
【分析】有一群人，有的人互为朋友，现在有一排椅子，将这些人安排在椅子上，要求不是朋友的两个人不能坐在一起，即可以将他俩隔开或者中间放个空的椅子。
已知K个人最多有K-1对朋友。问最少需要多少椅子。
树的最小路径覆盖模板题。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 200050;
const int M = 24005;
const int mod=1e9+7;
int n,m;
int T,cnt;
int head[N],ans[N],vis[N];
bool mark[N];
struct edge{int to,next;}edg[N*2];
void add(int u,int v)
{
edg[++cnt].to=v;edg[cnt].next=head[u];head[u]=cnt;
edg[++cnt].to=u;edg[cnt].next=head[v];head[v]=cnt;
}
void dfs(int x,int f)
{
ans[x]=vis[x]=1;
int tot=0;
for(int i=head[x];i!=-1;i=edg[i].next)
{
if(edg[i].to==f)continue;
dfs(edg[i].to,x);
ans[x]+=ans[edg[i].to];
if(!mark[edg[i].to])tot++;
}
if(tot>=2)ans[x]-=2,mark[x]=1;
else if(tot==1)ans[x]--;
}
int main()
{
int u,v;
scanf("%d",&T);
while(T--)
{
cnt=0;
met(head,-1);
met(ans,0);
met(mark,0);
met(vis,0);
scanf("%d%d",&n,&m);
while(m--)
{
scanf("%d%d",&u,&v);
add(u,v);
}
int anss=0,ret=0;;
for(int i=1;i<=n;i++){
if(!vis[i]){
ret++;
dfs(i,0);
vis[i]=1;
anss+=ans[i]-1;
}
}
printf("%d\n",anss+ret-1+n);
}
return 0;
}