树形dp

dp[i][j]=max(dp[i][j],dp[lch][k]+dp[rch][j-k])

#include
     
      using namespace std;int t, n, m, dp[1005][605];struct node {    int tim, pic; }tr[5005];inline int read() {    int red = 0, f_f = 1; char ch = getchar();    while(ch>'9' || ch <'0') {if(ch=='-') f_f = -1; ch = getchar();}    while(ch>='0' &&ch<='9') red = red*10+ch-'0', ch = getchar();    return red*f_f;}void dfs_read(int a) {    tr[a].tim = read() * 2, tr[a].pic = read();    if(!tr[a].pic) {        dfs_read(a << 1); dfs_read(a << 1| 1);    }}void dfs(int a, int r) {    if(dp[a][r] || !r) return；    if(tr[a].pic) {         dp[a][r] = min(tr[a].pic, (r - tr[a].tim) / 5);        return;    }    for (int i = 0; i <= r - tr[a].tim; i++) {         dfs(a << 1, i);         dfs(a << 1 | 1, r - tr[a].tim - i);         dp[a][r] = max(dp[a][r], dp[a << 1][i] + dp[a << 1 | 1][r - tr[a].tim - i]);    }}int main(){    memset(dp, 0, sizeof(dp));    t = read() - 1;     dfs_read(1);    dfs(1, t);    printf("%d\n", dp[1][t]);    return 0;}