并查集+01背包

题目给出人与人之间的关系，锁定一个数据结构那就是并查集，大概率和这个有关系，并且需要求解收益最大，想想带权并查集好像搞不定这个事。那当纯的并查集好像完成不了这个操作，那怎么办，既然是求最大值，二分？动规！打一个01背包去求解C中的价值最大。具体操作，把题目给出的存在关系的点直接连接在一起，这些有关系的点才可以花代价去认识。那么在01背包中第二维的部分必须要加个判断，是不是和1是同一个集合中，才可以进行01背包。好啦代码码好就行了，记得看清楚数据范围。

Code

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e4 + 7;
int fa[N];
int a[N], b[N];
int dp[505];

int find(int x) {
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}

int main() {
    int t = read();
    while (t--) {
        int  n = read(), m = read(), c = read();
        for (int i = 1; i <= n; ++i)    fa[i] = i;
        for (int i = 2; i <= n; ++i)
            a[i] = read(), b[i] = read();
        for (int i = 1; i <= m; ++i) {
            int x=read(), y=read();
            fa[find(x)] = find(y);
        }
        memset(dp, 0, sizeof(dp));
        for (int i = 2; i <= n; ++i)
            if (find(i) == find(1)) {
                for (int j = c; j >= a[i]; --j)
                    dp[j] = max(dp[j], dp[j - a[i]] + b[i]);
            }
        write(dp[c]),putchar(10);
    }
    return 0;
}