ASC7 Problem H. Oil Deal

题目大意

给个带边权图，求给定代价内最多删多少条边任然存在生成树。

简要题解

先做个最大生成树，然后从小往大删不在树上的边即可。

#include <bits/stdc++.h>
using namespace std;
namespace my_header {
#define pb push_back
#define mp make_pair
#define pir pair<int, int>
#define vec vector<int>
#define pc putchar
#define clr(t) memset(t, 0, sizeof t)
#define pse(t, v) memset(t, v, sizeof t)
#define bl puts("")
#define wn(x) wr(x), bl
#define ws(x) wr(x), pc(' ')
    const int INF = 0x3f3f3f3f;
    typedef long long LL;
    typedef double DB;
    inline char gchar() {
        char ret = getchar();
        for(; (ret == '
' || ret == '
' || ret == ' ') && ret != EOF; ret = getchar());
        return ret; }
    template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) {
        for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar());
        if (c == '-') { flg = -1; c = getchar(); }
        for(ret = 0; '0' <= c && c <= '9'; c = getchar())
            ret = ret * 10 + c - '0';
        ret = ret * flg; }
    inline int fr() { int t; fr(t); return t; }
    template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); }
    template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); }
    template<class T> inline char wr(T a, int b = 10, bool p = 1) {
        return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 
            (wr(a/b, b, 0), pc('0' + a % b)));
    }
    template<class T> inline void wt(T a) { wn(a); }
    template<class T> inline void wt(T a, T b) { ws(a), wn(b); }
    template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); }
    template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); }
    template<class T> inline T gcd(T a, T b) {
        return b == 0 ? a : gcd(b, a % b); }
    template<class T> inline T fpw(T b, T i, T _m, T r = 1) {
        for(; i; i >>= 1, b = b * b % _m)
            if(i & 1) r = r * b % _m;
        return r; }
};
using namespace my_header;

const int MAXN = 50000 + 100;
const int MAXM = 100000 + 100;
struct Edge {
    int u, v, w, id, blp;
    void read() {
        fr(u, v, w);
    }
    bool operator < (const Edge&rhs) const {
        return w < rhs.w;
    }
} e[MAXM];
LL n, m, s;

int fa[MAXN];
int findFa(int u) {
    return fa[u] == u ? u : fa[u] = findFa(fa[u]);
}

int main() {
#ifdef lol
    freopen("H.in", "r", stdin);
    freopen("H.out", "w", stdout);
#else
    freopen("oil.in", "r", stdin);
    freopen("oil.out", "w", stdout);
#endif
    fr(n, m, s);
    for (int i = 1; i <= m; ++i) {
        e[i].read();
        e[i].id = i;
    }
    sort(e + 1, e + m + 1);
    for (int i = 1; i <= n; ++i)
        fa[i] = i;
    for (int i = m; 0 < i; --i) {
        int u = findFa(e[i].u);
        int v = findFa(e[i].v);
        if (u != v) {
            fa[u] = v;
            e[i].blp = true;
        }
    }
    vector<int> res;
    for (int i = 1; i <= m; ++i) {
        if (s - e[i].w >= 0 && !e[i].blp) {
            s -= e[i].w;
            res.pb(e[i].id);
        }
    }
    wt(res.size());
    for (auto &&i : res) {
        ws(i);
    }
    bl;


    return 0;
}