C++ :

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
/* op1/op2 只让「被移动」的巧克力耐久 -1；op3 让整个盒子的巧克力耐久 -1。
   每次操作后把耐久为 0 的巧克力删掉。n,m<=1000, ci<=1000。 */
static vector<int> box[1005];
static void clean(vector<int> &v){
    vector<int>::iterator e = remove(v.begin(), v.end(), 0);
    v.erase(e, v.end());
}
int main(){
    int n, m;
    if(scanf("%d %d", &n, &m) != 2) return 0;
    for(int i = 1; i <= n; i++){
        int c;
        if(scanf("%d", &c) != 1) return 0;
        box[i].clear();
        for(int j = 0; j < c; j++){ int d; scanf("%d", &d); box[i].push_back(d); }
    }
    for(int t = 0; t < m; t++){
        int op;
        if(scanf("%d", &op) != 1) break;
        if(op == 1 || op == 2){
            int x, y;
            scanf("%d %d", &x, &y);
            int sz = (int)box[x].size();
            if(op == 1)
                for(int j = 0; j < sz; j++) box[y].push_back(box[x][j] - 1);
            else
                for(int j = sz - 1; j >= 0; j--) box[y].push_back(box[x][j] - 1);
            box[x].clear();
            clean(box[y]);
        } else {
            int x;
            scanf("%d", &x);
            for(size_t j = 0; j < box[x].size(); j++) box[x][j]--;
            clean(box[x]);
            reverse(box[x].begin(), box[x].end());
        }
    }
    for(int i = 1; i <= n; i++){
        printf("%d", (int)box[i].size());
        for(size_t j = 0; j < box[i].size(); j++) printf(" %d", box[i][j]);
        putchar('\n');
    }
    return 0;
}