SPOJ UOFTCG - Office Mates （树的最小路径覆盖）

UOFTCG - Office Mates

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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:

Output:

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 = ;

const int M = ;

const int mod=1e9+;

int n,m;

int T,cnt;

int head[N],ans[N],vis[N];

bool mark[N];

struct edge{int to,next;}edg[N*];

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]=;

    int tot=;

    for(int i=head[x];i!=-;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>=)ans[x]-=,mark[x]=;

    else if(tot==)ans[x]--;

}

int main()

{

    int u,v;

    scanf("%d",&T);

    while(T--)

    {

        cnt=;

        met(head,-);

        met(ans,);

        met(mark,);

        met(vis,);

        scanf("%d%d",&n,&m);

        while(m--)

        {

            scanf("%d%d",&u,&v);

            add(u,v);

        }

        int anss=,ret=;;

        for(int i=;i<=n;i++){

            if(!vis[i]){

                ret++;

                dfs(i,);

                vis[i]=;

                anss+=ans[i]-;

            }

        }

        printf("%d\n",anss+ret-+n);

    }

    return ;

}