1、题意：有一些点，点与点之间有保护关系，每个点都有一个权值，求能获得的最大值

2、分析：裸的最大权闭合图，用网络流进行求解，然后我们发现点与点之间的保护关系可能构成环，这样网络流是无法处理的，然后我们拓扑排序去掉那些不能获得的点。注意！！！！这里的环是不能用强连通来强行去掉的，因为——比如有一个点，他两端与它相连的点在环内，那么这个点你也去不掉

最大权闭合图模型：建立源点s和汇点t，将所有正权点连向s，容量为点权，将所有负权点连向t，容量为点权的相反数，原图中的边容量全部设成inf，跑一边最小割，用所有正权点的总和减一下就是答案了

#include <queue>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
#define M 500010
#define inf 214748364

inline int read(){
    char ch = getchar(); int x = 0, f = 1;
    while(ch < '0' || ch > '9'){
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while('0' <= ch && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

namespace dinic{
    struct Edge{
        int from, to, cap, flow, next;
    } G[M];
    int head[M], cur[M], tot;
    int vis[M], d[M];
    int s, t;

    inline void init(){
        tot = -1;
        memset(head, -1, sizeof(head));
    }

    inline void add(int u, int v, int w){
        G[++ tot] = (Edge){u, v, w, 0, head[u]};
        head[u] = tot;
        G[++ tot] = (Edge){v, u, 0, 0, head[v]};
        head[v] = tot;
    }

    inline bool BFS(){
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        vis[s] = 1;
        d[s] = 0;
        while(!Q.empty()){
            int x = Q.front(); Q.pop();
            for(int i = head[x]; i != -1; i = G[i].next){
                Edge& e = G[i];
                if(e.cap > e.flow && !vis[e.to]){
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    inline int DFS(int x, int a){
        if(x == t || a == 0) return a;
        int flow = 0, f;
        for(int &i = cur[x]; i != -1; i = G[i].next){
            Edge& e = G[i];
            if(d[e.to] == d[x] + 1 && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0){
                e.flow += f;
                G[i ^ 1].flow -= f;
                flow += f;
                a -= f;
                if(a == 0) break;
            }
        }
        return flow;
    }

    inline int ans(){
        int flow = 0;
        while(BFS()){
            for(int i = s; i <= t; i ++) cur[i] = head[i];
            flow += DFS(s, inf);
        }
        return flow;
    }
}

int val[M];

struct Edge{
    int u, v, next;
} G[M];
int head[M], tot;
int du[M], ok[M];
int n, m;

inline void addedge(int u, int v){
//  printf("%d %d\n", u, v);
    G[++ tot] = (Edge){u, v, head[u]};
    head[u] = tot;
    du[v] ++;
} 

inline void toposort(){
    queue<int> Q;
    for(int i = 1; i <= n * m; i ++) if(!du[i]){
        Q.push(i);
        ok[i] = 1;
    }
    while(!Q.empty()){
        int x = Q.front(); Q.pop();
        for(int i = head[x]; i != -1; i = G[i].next){
            du[G[i].v] --;
            if(!du[G[i].v]){
                Q.push(G[i].v);
                ok[G[i].v] = 1;
            }
            /*puts("fuck");
            printf("%d\n", i);*/
        }

    }
}

int main(){
    n = read(), m = read();
    #define Num(i, j) (i - 1) * m + j
    memset(head, -1, sizeof(head));
    for(int i = 1; i <= n; i ++){
        for(int j = 1; j <= m; j ++){
            int num = Num(i, j);
            val[num] = read(); int op = read();
            for(int k = 1; k <= op; k ++){
                int x = read(), y = read();
                x ++; y ++;
                addedge(num, Num(x, y));
            }
            if(j != m) addedge(num + 1, num);
        }
    }
    toposort();
    //puts("fuck");
    dinic::s = 0; dinic::t = n * m + 1;
    dinic::init();
    int res = 0;
    for(int i = 1; i <= n * m; i ++) if(ok[i]){
        if(val[i] > 0) res += val[i], dinic::add(dinic::s, i, val[i]);
        else dinic::add(i, dinic::t, -val[i]);
        for(int j = head[i]; j != -1; j = G[j].next){
            if(ok[G[j].v]){
                dinic::add(G[j].v, i, inf);
            }
        }
    }

    printf("%d\n", res - dinic::ans());
    return 0;
}