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。
1 #include <bits/stdc++.h> 2 using namespace std; 3 namespace my_header { 4 #define pb push_back 5 #define mp make_pair 6 #define pir pair<int, int> 7 #define vec vector<int> 8 #define pc putchar 9 #define clr(t) memset(t, 0, sizeof t) 10 #define pse(t, v) memset(t, v, sizeof t) 11 #define bl puts("") 12 #define wn(x) wr(x), bl 13 #define ws(x) wr(x), pc(' ') 14 const int INF = 0x3f3f3f3f; 15 typedef long long LL; 16 typedef double DB; 17 inline char gchar() { 18 char ret = getchar(); 19 for(; (ret == ' ' || ret == ' ' || ret == ' ') && ret != EOF; ret = getchar()); 20 return ret; } 21 template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) { 22 for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar()); 23 if (c == '-') { flg = -1; c = getchar(); } 24 for(ret = 0; '0' <= c && c <= '9'; c = getchar()) 25 ret = ret * 10 + c - '0'; 26 ret = ret * flg; } 27 inline int fr() { int t; fr(t); return t; } 28 template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); } 29 template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); } 30 template<class T> inline char wr(T a, int b = 10, bool p = 1) { 31 return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 32 (wr(a/b, b, 0), pc('0' + a % b))); 33 } 34 template<class T> inline void wt(T a) { wn(a); } 35 template<class T> inline void wt(T a, T b) { ws(a), wn(b); } 36 template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); } 37 template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); } 38 template<class T> inline T gcd(T a, T b) { 39 return b == 0 ? a : gcd(b, a % b); } 40 template<class T> inline T fpw(T b, T i, T _m, T r = 1) { 41 for(; i; i >>= 1, b = b * b % _m) 42 if(i & 1) r = r * b % _m; 43 return r; } 44 }; 45 using namespace my_header; 46 47 const int MAXN = 50000 + 100; 48 const int MAXM = 100000 + 100; 49 struct Edge { 50 int u, v, w, id, blp; 51 void read() { 52 fr(u, v, w); 53 } 54 bool operator < (const Edge&rhs) const { 55 return w < rhs.w; 56 } 57 } e[MAXM]; 58 LL n, m, s; 59 60 int fa[MAXN]; 61 int findFa(int u) { 62 return fa[u] == u ? u : fa[u] = findFa(fa[u]); 63 } 64 65 int main() { 66 #ifdef lol 67 freopen("H.in", "r", stdin); 68 freopen("H.out", "w", stdout); 69 #else 70 freopen("oil.in", "r", stdin); 71 freopen("oil.out", "w", stdout); 72 #endif 73 fr(n, m, s); 74 for (int i = 1; i <= m; ++i) { 75 e[i].read(); 76 e[i].id = i; 77 } 78 sort(e + 1, e + m + 1); 79 for (int i = 1; i <= n; ++i) 80 fa[i] = i; 81 for (int i = m; 0 < i; --i) { 82 int u = findFa(e[i].u); 83 int v = findFa(e[i].v); 84 if (u != v) { 85 fa[u] = v; 86 e[i].blp = true; 87 } 88 } 89 vector<int> res; 90 for (int i = 1; i <= m; ++i) { 91 if (s - e[i].w >= 0 && !e[i].blp) { 92 s -= e[i].w; 93 res.pb(e[i].id); 94 } 95 } 96 wt(res.size()); 97 for (auto &&i : res) { 98 ws(i); 99 } 100 bl; 101 102 103 return 0; 104 }