比较奇怪的树形背包
首先需要处理读入的问题 这题史诗递归读入
然后递归读入就不用建图
这题特点是只有叶子有价值 所以背包就不太有用
坑点就是这个人进去还得出来...
而且不能把时间都用完(导致75)
Time cost: 35min
Code:
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 #include<cmath>
5 #include<queue>
6 #define ms(a,b) memset(a,b,sizeof a)
7 #define rep(i,a,n) for(int i = a;i <= n;i++)
8 #define per(i,n,a) for(int i = n;i >= a;i--)
9 #define inf 2147483647
10 using namespace std;
11 typedef long long ll;
12 typedef double D;
13 #define eps 1e-8
14 ll read() {
15 ll as = 0,fu = 1;
16 char c = getchar();
17 while(c < '0' || c > '9') {
18 if(c == '-') fu = -1;
19 c = getchar();
20 }
21 while(c >= '0' && c <= '9') {
22 as = as * 10 + c - '0';
23 c = getchar();
24 }
25 return as * fu;
26 }
27 //head
28 const int N = 1005;
29 int n,V;
30 struct node {
31 int val,cst;
32 }p[N<<4];
33 int dp[N][N];
34
35 #define ls x<<1
36 #define rs x<<1|1
37 void input(int x) {
38 p[x].cst = read() << 1,p[x].val = read();
39 if(!p[x].val) input(ls),input(rs);
40 }
41
42 int dfs(int x,int tot) {
43 if(!tot) return 0;
44 if(dp[x][tot]) return dp[x][tot];
45 //sn
46 if(p[x].val) return dp[x][tot] = min(p[x].val,(tot - p[x].cst) / 5);
47 //pa
48 rep(k,0,tot - p[x].cst)
49 dp[x][tot] = max(dp[x][tot],dfs(ls,k) + dfs(rs,tot - p[x].cst - k));
50 return dp[x][tot];
51 }
52
53 int main() {
54 int V = read() - 1;
55 input(1);
56 printf("%d
",dfs(1,V));
57 return 0;
58 }