「JSOI2016」最佳团体

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
 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++)
22               f[u][j] = max(f[u][j], f[v][k] + f[u][j - k]);
23     }
24 }
25 
26 bool check(double x) {
27     memset(f, -0x3f, sizeof(f));
28     for(int i = 0; i <= n; i++) f[i][0] = 0;
29     for(int i = 1; i <= n; i++) w[i] = (double) p[i] - s[i]*x;
30     dfs(0);
31     return f[0][m] >= 0;
32 }
33 
34 void add(int u, int v) {
35     e[cnt].to = v;
36     e[cnt].next = first[u];
37     first[u] = cnt++;
38 }
39 
40 int main() {
41     memset(first, -1, sizeof(first));
42     cin>>m>>n;
43     m++;
44     for(int i = 1; i <= n; i++) {
45         scanf("%d%d%d", &s[i], &p[i], &fa[i]);
46         add(fa[i], i);
47    }
48    while(r - l > eps) {
49        mid = (r + l) / 2;
50        dfs(0);
51        if(check(mid)) l = mid + eps, ans = mid;
52        else r = mid - eps;
53    }
54    printf("%.3f\n", ans);
55 }

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