牛客练习赛43C Tachibana Kanade Loves Review

题目链接：https://ac.nowcoder.com/acm/contest/548/C

题目大意：

　　略

分析：

　　这是一道求最小生成树的模板题，对于给定的知识点，可以假想一个已经学会的 0 号知识点，那么所有知识点之间就都可以用带权值的边关联起来了，如果某个知识点已经学完，那么从 0 号知识点学习那个知识点的边权值就设为 0 即可。

　　PS:我当初做的时候数据并不强，现在提交过去AC的代码居然Wa，优化了一下才过（不要用结构体，尽量避免函数调用）。

代码如下：

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl

#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())

#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))

#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef long long LL;
const int maxN = 1e6 + 7;

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

struct Vertex;
struct Edge;

int n, m, k, ans; 
LL t;

int f[maxN], sz[maxN];

inline void Find(int &x){
    while (x != f[x]) x = f[x] = f[f[x]];
} 

inline void Union(int x, int y){
    if (sz[x] > sz[y]) swap(x, y);
    f[x] = y;
}


struct Edge{
    int weight;
    int from, to;
    
    Edge(){}
    
    Edge(int w, int f, int t) : weight(w), from(f), to(t){}
    bool operator < (const Edge &x) const {
        return weight < x.weight;
    }
    
    void print() const{
        printf("Edge:
");
        printf("    ");prln(weight);
        printf("    ");prln(from);
        printf("    ");prln(to);
    }
};

Edge E[maxN*10];
int elen = 0;

// 返回值为最小生成树的权值总和 
inline int kruskal(int n){
    rep(i, n+1){
        f[i] = i;
    }

    int cnt = 0, ret = 0;

    sort(E + 1, E + elen + 1);

    For(i, 1, elen){
        int x = E[i].from, y = E[i].to;

        Find(x);
        Find(y);
        if(x == y) continue;
        else {
            Union(x, y);
            ++cnt;
            ret += E[i].weight;
            if(ret > t) return ret;
        }
    }
    return ret;
}

int main(){
    scanf("%d%d%d%d", &n, &m, &k, &t);
    // 假想一个0号知识点，并且已经会了
    For(i, 1, n) E[++elen] = Edge(ri(), 0, i);
    For(i, 1, k) E[ri()].weight = 0;
    For(i, 1, m){
        int x = ri(), y = ri(), h = ri();
        E[++elen] = Edge(h, x, y);
    }
    // 求最小生成树
    ans = kruskal(n);
    
    if(ans > t) printf("No
");
    else printf("Yes
");
    return 0;
}