[AcWing 1252] 搭配购买

并查集 + 01背包

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int n, m, k;
int p[N];
int v[N], w[N];
int f[N];

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

void solve()
{
    cin >> n >> k >> m;
    for (int i = 1; i <= n; i ++)
        p[i] = i;
    for (int i = 1; i <= n; i ++)
        cin >> v[i] >> w[i];
    while (k --) {
        int a, b;
        cin >> a >> b;
        int pa = find(a), pb = find(b);
        if (pa != pb) {
            v[pb] += v[pa];
            w[pb] += w[pa];
            p[pa] = pb;
        }
    }
    for (int i = 1; i <= n; i ++)
        if (p[i] == i)
            for (int j = m; j >= v[i]; j --)
                f[j] = max(f[j], f[j - v[i]] + w[i]);
    cout << f[m] << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    solve();

    return 0;
}

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1. 把每个连通块看作是一个物品，转换成 01 背包问题
2. 每个连通块的体积和价值都存到并查集的根节点