BZOJ 1040 骑士(环套树DP)

时间：2024-11-12 23:04:56

如果m=n-1，显然这就是一个经典的树形dp。

现在是m=n，这是一个环套树森林，破掉这个环后，就成了一个树，那么这条破开的边连接的两个顶点不能同时选择。我们可以对这两个点进行两次树形DP根不选的情况。

那么答案就是每个森林的max()之和。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF (LL)<<

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

struct Edge{int p, next, flag;}edge[N<<];

int head[N], cnt=, node[N], res[], p;

bool vis[N];

LL dp[][N][];

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}

void dfs(int x, int fa)

{

    vis[x]=true;

    for (int i=head[x]; i; i=edge[i].next) {

        int v=edge[i].p;

        if (v==fa) continue;

        if (vis[v]==true) {

            if (p==) res[]=x, res[]=v, p=, edge[i].flag=edge[i^].flag=;

            continue;

        }

        dfs(v,x);

    }

}

void dfs_dp(int x, int fa, int flag)

{

    for (int i=head[x]; i; i=edge[i].next) {

        int v=edge[i].p;

        if (v==fa||edge[i].flag) continue;

        dfs_dp(v,x,flag);

        dp[flag][x][]+=max(dp[flag][v][],dp[flag][v][]);

        dp[flag][x][]+=dp[flag][v][];

    }

    dp[flag][x][]+=node[x];

}

int main ()

{

    int n, u, w;

    LL ans=;

    scanf("%d",&n);

    FOR(i,,n) scanf("%d%d",&w,&u), add_edge(u,i), add_edge(i,u), node[i]=w;

    FOR(i,,n) {

        if (vis[i]) continue;

        p=; dfs(i,);

        dfs_dp(res[],,); dfs_dp(res[],,);

        ans+=max(dp[][res[]][],dp[][res[]][]);

    }

    printf("%lld\n",ans);

    return ;

}

相关文章