bzoj 1791 DP

　　首先对于一棵树我们可以tree_dp来解决这个问题，那么对于环上每个点为根的树我们可以求出这个树的一端为根的最长链，并且在tree_dp的过程中更新答案。那么我们对于环，从某个点断开，破环为链，然后再用DP来解决这个问题。

　　备注：很久之前的一道题，刚转的c++，然后T了，也懒得改了。

/**************************************************************
    Problem: 1791
    User: BLADEVIL
    Language: C++
    Result: Time_Limit_Exceed
****************************************************************/

//By BLADEVIL
#include <cstdio>
#include <cstring>
#define maxn 1000010
#define LL long long 

using namespace std; 

LL n,time,l;
LL other[maxn<<],last[maxn],pre[maxn<<],dfn[maxn],low[maxn],vis[maxn],que[maxn],a[maxn],xx[maxn];
LL ans;
LL len[maxn<<],max1[maxn],max2[maxn],sum[maxn],w[maxn],yy[maxn],maxlen[maxn];
LL save; 

void getmin(LL &x,LL y)
{if (y<x) x=y;} 

void connect(LL x,LL y,LL z) {
    pre[++l]=last[x];
    last[x]=l;
    other[l]=y;
    len[l]=z;
    //if (y>x) printf("%d %d %lld\n",x,y,z);
} 

void dfs(LL x,LL fa) {
    dfn[x]=low[x]=++time;
    for (LL q=last[x];q;q=pre[q]) {
        if (other[q]==fa) continue;
        if (!low[other[q]]) {
            dfs(other[q],x);
            getmin(low[x],low[other[q]]);
        } else getmin(low[x],dfn[other[q]]);
    }
    if (low[x]!=dfn[x]) save=low[x];
} 

void dp(LL x) {
    memset(que,,sizeof que);
    LL h=,t=1ll;
    que[]=x; vis[x]=1ll;
    while (h<t) {
        LL cur=que[++h];
        for (LL q=last[cur];q;q=pre[q]) {
            if (low[other[q]]==low[x]) continue;
            if (vis[other[q]]) continue;
            vis[other[q]]=vis[cur]+1ll;
            que[++t]=other[q];
        }
    }
    //for (LL i=1;i<=t;i++) printf("%d ",que[i]); printf("\n");
    for (LL i=t;i;i--)
        for (LL q=last[que[i]];q;q=pre[q]) {
            if (low[other[q]]==low[x]) continue;
            if (vis[other[q]]!=vis[que[i]]+) continue;
            //printf(" %d %d\n",que[i],other[q]);
            if (max1[other[q]]+len[q]>max1[que[i]])
                max2[que[i]]=max1[que[i]],max1[que[i]]=max1[other[q]]+len[q]; else
            if (max1[other[q]]+len[q]>max2[que[i]]) max2[que[i]]=max1[other[q]]+len[q];
            if (max1[other[q]]+max2[other[q]]>maxlen[que[i]])
                maxlen[que[i]]=max1[other[q]]+max2[other[q]];
            if (max1[que[i]]+max2[que[i]]>maxlen[que[i]])
                maxlen[que[i]]=max1[que[i]]+max2[que[i]];
            if (maxlen[que[i]]<maxlen[other[q]]) maxlen[que[i]]=maxlen[other[q]];
        }
    //for (LL i=1;i<=t;i++) printf("%d %lld %lld %lld\n",que[i],max1[que[i]],max2[que[i]],maxlen[que[i]]);
} 

void solve(LL x) {
    LL now=0ll;
    dfs(x,-);
    //printf("%d",save);
    if (save)
        for (LL i=;i<=n;i++) if (low[i]==save) x=i;
    save=;
    /*if (xx[xx[x]]==x)
    {
        ans+=(yy[x]>yy[xx[x]])?yy[x]:yy[xx[x]];
        return;
    }*/
    LL cur;
    for (LL i=;i<=n;i++) if (low[i]==low[x]) dp(i),cur=i;
    //printf(" %lld\n",max1[x]);
    LL t=; a[t]=cur;
    while () {
        for (LL q=last[a[t]];q;q=pre[q]) {
            now=(maxlen[a[t]]>now)?maxlen[a[t]]:now;
            if (low[other[q]]!=low[x]) continue;
            if (other[q]==a[t-]) continue;
            a[++t]=other[q]; sum[t]=len[q]; break;
        }
        if (a[t]==cur) break;
    }
    //printf(" %d ",x);
    //for (LL i=1;i<=t;i++) printf(" %lld %d %d\n",max1[a[i]],a[i],sum[i]);
    t--;
    for (LL i=;i<=t;i++) a[i+t]=a[i],sum[i+t]=sum[i];
    t*=;
    //for (LL i=1;i<=t;i++) printf(" %lld %d %d\n",max1[a[i]],a[i],sum[i]);
    for (LL i=;i<=t;i++) sum[i]+=sum[i-];
    LL len=t>>1ll;
    memset(que,,sizeof que);
    LL l=,r=; que[]=;
    for (LL i=;i<=t;i++) {
        if (i-que[l]+>len) l++;
        w[i]=max1[a[que[l]]]+max1[a[i]]+sum[i]-sum[que[l]];
        //printf("w[i]=%d",w[i]); printf(" %d %d\n",i,que[l]);
        while (l<=r&&(max1[a[i]]-sum[i]>max1[a[que[r]]]-sum[que[r]])) r--;
        que[++r]=i;
        //for (LL i=l;i<=r;i++) printf("|%d ",que[i]); printf("\n");
    }
    for (LL i=;i<=t;i++) now=(w[i]>now)?w[i]:now;
    //printf(" %lld ",ans);
    ans+=now;
} 

int main() {
    scanf("%d",&n);
    for (LL i=;i<=n;i++) scanf("%d%lld",&xx[i],&yy[i]);
    for (LL i=;i<=n;i++) {
        if (xx[i]==i) continue;
        if (xx[xx[i]]==i&&xx[i]>i) yy[i]=(yy[xx[i]]>yy[i])?yy[xx[i]]:yy[i],yy[xx[i]]=-1ll;
    }
    for (LL i=;i<=n;i++) if (yy[i]!=-) connect(i,xx[i],yy[i]),connect(xx[i],i,yy[i]);
    for (LL i=;i<=n;i++) if (!low[i]) solve(i);
    //for (LL i=1;i<=n;i++) printf(" %d %d %d\n",i,low[i],dfn[i]);
    printf("%lld\n",ans);
    return ;
}