题目链接：https://www.luogu.com.cn/problem/P4043
题目大意：
图片说明

思路：就是限制流量[1, inf]。有个坑点，就是汇点可以是任意点。不一定是剧情结束点。
例如：
图片说明 

#pragma GCC optimize(3, "Ofast", "inline")
#include<bits/stdc++.h>
#define re register
using namespace std;

const int maxn = 500 + 10;
const int maxm=1e6+10;
const int inf=1<<30;

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
char buf[1<<20],*p1=buf,*p2=buf;
inline int read() {
    int f=0,fu=1;
    char c=getchar();
    while(c<'0'||c>'9') {
        if(c=='-')
            fu=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9') {
        f=(f<<3)+(f<<1)+c-48;
        c=getchar();
    }
    return f*fu;
}

struct po {
    int to,dis,nxt,w;
} edge[maxm<<1];

struct max_folw {
    int dis[maxn];
    int vis[maxn];
    int head[maxn];
    int d[maxn];
    int n,s,t,cut=-1;
    int ans=0, fy=0;

    void init(int N, int S, int T) {
        ans=0, fy=0;
        cut=-1;
        n=N, s=S, t=T;
        memset(head, -1, sizeof(head));
        memset(d, 0, sizeof(d));
    }

    void add_edge(int from,int to,int w,int dis) {
        edge[++cut].nxt=head[from];
        edge[cut].to=to;
        edge[cut].w=w;
        edge[cut].dis=dis;
        head[from]=cut;
    }
    void add(int from,int to,int w,int dis) {
        add_edge(from,to,w,dis);
        add_edge(to,from,0,-dis);
    }

    void addlr(int from,int to,int w,int l, int r){
        add(from,to,r-l,w);
        d[to]+=l, d[from]-=l;
        fy+=l*w;
    }

    bool spfa() {
        memset(vis,0,sizeof(vis));
        for(re int i=0; i<=n; i++)
            dis[i]=inf;
        dis[t]=0;
        vis[t]=1;
        deque<int> q;
        q.push_back(t);
        while(!q.empty()) {
            int u=q.front();
            vis[u]=0;
            q.pop_front();
            for(re int i=head[u]; i!=-1; i=edge[i].nxt) {
                int v=edge[i].to;
                if(edge[i^1].w>0&&dis[v]>dis[u]-edge[i].dis) {
                    dis[v]=dis[u]-edge[i].dis;
                    if(!vis[v]) {
                        vis[v]=1;
                        if(!q.empty()&&dis[v]<dis[q.front()])
                            q.push_front(v);
                        else
                            q.push_back(v);
                    }
                }
            }
        }
        return dis[s]<inf;
    }
    int dfs(int u,int low) {
        if(u==t) {
            vis[t]=1;
            return low;
        }
        int diss=0;
        vis[u]=1;
        for(re int i=head[u]; i!=-1; i=edge[i].nxt) {
            int v=edge[i].to;
            if(!vis[v]&&edge[i].w!=0&&dis[u]-edge[i].dis==dis[v]) {
                int check=dfs(v,min(edge[i].w,low));
                if(check>0) {
                    fy+=check*edge[i].dis;
                    edge[i].w-=check;
                    edge[i^1].w+=check;
                    low-=check;
                    diss+=check;
                    if(low==0)
                        break;
                }
            }
        }
        return diss;
    }
    void max_flow() {

        for(int i=0; i<=n; i++){
            if(d[i]>0){
                add(s, i, d[i], 0);
            }
            if(d[i]<0){
                add(i, t, -d[i], 0);
            }
        }

        while(spfa()) {
            vis[t]=1;
            while(vis[t]) {
                memset(vis,0,sizeof(vis));
                ans+=dfs(s,inf);
            }
        }
    }
} flow;

int d[maxn];
int main() {


    int n; scanf("%d", &n);
    int s=1, t=n+1, ss=n+2, tt=n+3;//
    flow.init(tt+5, ss, tt);

    for(int i=1; i<=n; i++){
        int k; scanf("%d", &k);
        int from=i, to, fy, l=1, r=inf;
        flow.add(i, t, inf, 0);
        while(k--){
            scanf("%d%d", &to, &fy);
            flow.addlr(i, to, fy, l, r);
        }
    }

    flow.add(t, s, inf, 0);
    flow.max_flow();
    cout<<flow.fy<<endl;

    return 0;
}