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本文详细介绍了一个基于图论的算法实现过程，包括初始化图结构、深度优先搜索算法的应用以及动态规划求解特定问题的方法。该算法旨在解决图中节点间特定类型的路径寻找问题，并通过实例演示了如何使用该算法进行有效求解。

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重做这道题

http://blog.youkuaiyun.com/woshi250hua/article/details/7632785

http://blog.youkuaiyun.com/shuangde800/article/details/10150305

http://blog.youkuaiyun.com/alps233/article/details/51190997

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define MAXN 500+5
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue
#define INF 0x3f3f3f3f

#define ls (rt<<1)
#define rs (rt<<1|1)

int n,m;

int head[MAXN],vis[MAXN],ptr=1;

int dp[MAXN][MAXN],num[MAXN];

struct node{int y,next,val;}tree[MAXN<<2];

void init()
{
    mem(head,-1);
    mem(vis,0);
    mem(dp,INF);
    mem(num,0);
    ptr = 1;
}

void add(int son,int fa)
{
    tree[ptr].y=son;
    tree[ptr].next=head[fa];
    head[fa]=ptr++;
}

void dfs(int rt)
{
    num[rt] = vis[rt] = 1;

    int tot = 0;

    for(int i = head[rt];i!=-1;i=tree[i].next)
    {
        int y = tree[i].y;
        if(vis[y]) continue;
        dfs(y);
        tot++;
        num[rt]+=num[y];
    }

    dp[rt][1] = tot;

    for(int i = head[rt];i!=-1;i=tree[i].next)
    {
        int y = tree[i].y;
        for(int j = num[rt];j>1;j--)
        {
            for(int k = 1;k<j;k++)
            {
                if(dp[rt][j-k]!=INF && dp[y][k]!=INF);
                    dp[rt][j] = min(dp[rt][j],dp[rt][j-k]+dp[y][k]-1);
            }
        }
    }
}


int main()
{
    int i,j,k=1;
    while(~sf("%d%d",&n,&m))
    {
        init();
        for(i=1;i<n;i++)
        {
            int x,y;
            sf("%d%d",&x,&y);
            add(x,y);
            add(y,x);
        }
        dfs(1);
        int ans = INF;
        for(i=1;i<=n;i++)
        {
            if(i==1) ans = min(dp[i][m],ans);
            else ans = min(dp[i][m]+1,ans);
        }
        pf("%d\n",ans);
    }
}