标签:dfs main ace can 简单 const for check bool
01分数规划
显然可以二分最大比值x,来验证是否可行
记当前比值为x,总战斗值为P与总招募费用为S
则 P - x*S >= 0
设 wi = pi - x*si
即 w1 + w2 + ... + wk >= 0
就转化为选k个节点,它们的w值非负,树上简单地dp一下就可求得
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int N = 2505; 5 6 double l, r = 1e4, mid, f[N][N], w[N], ans, eps = 1e-5; 7 int siz[N], fa[N], p[N], s[N]; 8 int n, m, first[N], cnt; 9 struct Edge { 10 int to, next; 11 } e[N]; 12 13 void dfs(int u) { 14 siz[u] = 1; 15 f[u][1] = w[u]; 16 for (int i = first[u]; i != -1; i = e[i].next) { 17 int v = e[i].to; 18 dfs(v); 19 siz[u] += siz[v]; 20 for (int j = min(siz[u], m); j >= 2; j--) 21 for (int k = 1; k <= min(j - 1, siz[v]); k++) f[u][j] = max(f[u][j], f[v][k] + f[u][j - k]); 22 } 23 } 24 25 bool check(double x) { 26 memset(f, -0x3f, sizeof(f)); 27 for (int i = 0; i <= n; i++) f[i][0] = 0; 28 for (int i = 1; i <= n; i++) w[i] = (double)p[i] - s[i] * x; 29 dfs(0); 30 return f[0][m] >= 0; 31 } 32 33 void add(int u, int v) { 34 e[cnt].to = v; 35 e[cnt].next = first[u]; 36 first[u] = cnt++; 37 } 38 39 int main() { 40 memset(first, -1, sizeof(first)); 41 cin >> m >> n; 42 m++; 43 for (int i = 1; i <= n; i++) { 44 scanf("%d%d%d", &s[i], &p[i], &fa[i]); 45 add(fa[i], i); 46 } 47 while (r - l > eps) { 48 mid = (r + l) / 2; 49 dfs(0); 50 if (check(mid)) 51 l = mid + eps, ans = mid; 52 else 53 r = mid - eps; 54 } 55 printf("%.3f\n", ans); 56 }
标签:dfs main ace can 简单 const for check bool
原文地址:https://www.cnblogs.com/ympc2005/p/12307225.html