本题大意：绑定销售求最大价值
对于绑定的价值可以采用并查集进行维护每一个’大物品’的v和w，完成操作后做一次01背包问题即可

import java.util.*;

public class Main {
    static final int N = 10010;
    static int[] f = new int[N], p = new int[N];
    static int[] v = new int[N], w = new int[N];
    static int n, m, vol;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        vol = sc.nextInt();

        for (int i = 1; i <= n; i++) p[i] = i;

        for (int i = 1; i <= n; i++) {
            v[i] = sc.nextInt();
            w[i] = sc.nextInt();
        }

        //并查集操作
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt(), b = sc.nextInt();
            int pa = find(a), pb = find(b);
            if (pa != pb) {//不能在同一个集合中
                v[pb] += v[pa];数值存储在根节点中
                w[pb] += w[pa];
                p[pa] = pb;//将pa连接到pb上
            }
        }

        //01背包
        for (int i = 1; i <= n; i++)
            if (i == p[i])//需要时根节点的物品才有意义
                for (int j = vol; j >= v[i]; j--)
                    f[j] = Math.max(f[j], f[j - v[i]] + w[i]);

        System.out.println(f[vol]);
    }

    static int find(int u) {
        if (u != p[u]) p[u] = find(p[u]);
        return p[u];
    }
}