LuoGuP2607:[ZJOI2008]骑士

Pre

之前做过这一题，方法比较麻烦。

Solution

基环树乱搞。

简便的是把环断开，以两个断点为根进行 (DP) 这样就可以算出答案。

注意到 (STD) 代码里面是有一句

dp[v][1] = -1e9;

这样就强制的选了一个，不选另外一个。

Code

Code(!std)

#include<cstdio>
#include<cstring>
#include<string>
#include<iostream>
#include<unistd.h>
#include<utility>
#include<map>
#include<limits.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rint register int
#define ll long long
#define I inline
#define ill I ll
#define iint I int
#define ivoid I void
#define ifloat I float
#define idouble I double
#define ibool bool
#define ipair I pair
#define xx first
#define yy second
using namespace std;
int mod;
ill qpow(ll m,ll n)
{
    ll base=m,tot=1;
    while(n)
    {
        if(n%2==1){
            tot*=base;
            tot%=mod;
        }
        base*=base;
        base%=mod;
        n/=2;
    }
    return tot%mod;
}
struct _in{
    const _in&operator,(rint&a)const{
        a=0;
        rint f=1;
        char k=getchar();
        while(k>'9'||k<'0'){if(k=='-')f=-1;k=getchar();}
        while(k>='0'&&k<='9'){a=a*10+k-'0';k=getchar();}
        a*=f;
        return*this;
    }
}in;

const int maxn=1005000;

//链式前向星
int h[maxn],fr[maxn],to[maxn],tot;
ivoid add(rint,rint);
//End

int n;
int fa[maxn],dep[maxn];
ll data[maxn];
int sign;
ll ans=0;
bool ed[maxn];
ll f[maxn][3];

void dp(rint,rint);

int main()
{
    //freopen("data.in","r",stdin);
    in,n;
    for(rint i=1;i<=n;i++){
        scanf("%lld",&data[i]);
        in,fa[i];
        add(fa[i],i);
    }
    for(rint i=1;i<=n;i++){
        if(!ed[i]){
            rint root=i;
            ed[root]=1;
            while(!ed[fa[root]]){
                root=fa[root];
                ed[root]=1;
            }
            dp(root,root);
            ll tmax=f[root][0];
            // for(rint i=1;i<=n;i++) printf("%d %d
",f[i][0],f[i][1]);
            if(root!=fa[sign]){
               pair<ll,ll>now,pre;
               now=make_pair(f[sign][0],f[sign][0]);
               pre=make_pair(f[sign][0],f[sign][1]);
               rint Now=fa[sign];
               while(1)
               {
                    pair<ll,ll> tmp=make_pair(f[Now][0],f[Now][1]);
                    f[Now][0]-=max(pre.xx,pre.yy);
                    f[Now][0]+=max(now.xx,now.yy);
                    f[Now][1]-=pre.xx;
                    f[Now][1]+=now.xx;
                    pre=tmp;
                    now=make_pair(f[Now][0],f[Now][1]);
                    if(Now==root) break;
                    Now=fa[Now];
               }
            }
            ans+=max(f[root][1],max(tmax,f[root][0]));
        }
    }
    printf("%lld
",ans);
    return 0;
}


void dp(rint pos,rint root)
{
    f[pos][1]=data[pos];
    dep[pos]=dep[fa[pos]]+1;
    ed[pos]=1;
    for(rint i=h[pos];i;i=fr[i]){
        if(to[i]!=root){
            dp(to[i],root);
            f[pos][1]+=f[to[i]][0];
            f[pos][0]+=max(f[to[i]][0],f[to[i]][1]);
        }
        else{
            sign=pos;
        }
    }
}

ivoid add(rint u,rint v)
{
    tot++;
    fr[tot]=h[u];
    to[tot]=v;
    h[u]=tot;
}

Code(std)

#include <bits/stdc++.h>
#define LL long long
#define maxn 1000010
using namespace std;
struct Edge{
    int to, next;
}edge[maxn << 1];
int num, head[maxn], a[maxn], f[maxn], vis[maxn], n;
LL dp[maxn][2], ans;

inline int read(){
    int s=  0,w = 1;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
    for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
    return s * w;
}

void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}, head[x] = num; }

void dfs(int u, int x){
    vis[u] = 1;
    dp[u][0] = 0, dp[u][1] = a[u];
    for (int i = head[u]; i; i = edge[i].next){
        int v = edge[i].to;
        if (v != x){
            dfs(v, x);
            dp[u][0] += max(dp[v][0], dp[v][1]);
            dp[u][1] += dp[v][0];
        } else dp[v][1] = -1e9;
    }
}

void solve(int u){ 
    while (!vis[u]) vis[u] = 1, u = f[u];
    dfs(u, u);
    LL sum = max(dp[u][0], dp[u][1]);
    u = f[u];
    dfs(u, u);
    ans += max(sum, max(dp[u][0], dp[u][1]));
}

int main(){
    n = read();
    for (int i = 1; i <= n; ++i){
        a[i] = read(), f[i] = read();
        addedge(f[i], i);
    }
    for (int i = 1; i <= n; ++i)
        if (!vis[i]) solve(i);
    printf("%lld
", ans);
    return 0;
}

Conclusion

有时候考虑一些简单的做法会有很大的好处。