题目地址

Link

题解

虚点这种东西还是没有掌握好啊。
考虑建一个虚点，向已经学会的东西连一条边权为0的边，关系正常连边，单独学的从虚点连一条边过去。
然后做一遍最小生成树就得到答案了。
这题略卡常，上个快读稳一点。

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <deque>
#include <ctime>
#include <map>
#include <set>

#define ll long long
#define inf 0x3f3f3f3f
#define il inline

namespace io {

#define in(a) a = read()
#define out(a) write(a)
#define outn(a) out(a), putchar('\n')

#define I_int ll
inline I_int read() {
    I_int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
char F[200];
inline void write(I_int x) {
    if (x == 0) return (void) (putchar('0'));
    I_int 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]);
}
#undef I_int

}
using namespace io;

using namespace std;

#define N 8000010

int n, m, k, t, a[N], f[N];
struct edge {
    int u, v, w;
}e[N];

bool operator < (edge a, edge b) {
    return a.w < b.w;
}

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

int main() {
    in(n), in(m), in(k), in(t);
    int cnt = 0;
    for(int i = 1; i <= n; ++i) {
        int x = read();
        e[++cnt] = (edge) {i, n + 1, x};
    }
    ++n; 
    for(int i = 1; i <= n; ++i) f[i] = i;
    for(int i = 1; i <= k; ++i) {
        int x = read();
        e[++cnt] = (edge) {x, n, 0};
    }
    for(int i = 1; i <= m; ++i) {
        int u = read(), v = read(), w = read();
        e[++cnt] = (edge) {u, v, w};
    }
    sort(e+1,e+cnt+1);
    ll ans = 0;
    for(int i = 1; i <= cnt; ++i) {
        int x = find(e[i].u), y = find(e[i].v);
        if(x != y) {
            f[y] = x;
            ans += e[i].w;
        }
    }
    if(ans > t) puts("No");
    else puts("Yes");
//  outn(ans);
}