题目链接：
题意：有两类装备，高级装备A和基础装备B。现在有m的钱。每种B有一个单价和可以购买的数量上限。每个Ai可以由Ci种其他物品合成，给出Ci种其他物品每种需要的数量。每个装备有一个贡献值。求最大的贡献值。已知物品的合成路线是一个严格的树模型。即有一种物品不会合成其他任意物品，其余物品都会仅仅可用作合成另外一种物品。
思路：树形DP。 f[i][j][k]表示以第i种物品为根的子树花费j，向上层提供k个i所得到的的最大贡献值。
#include <iostream>
#include <cstdio>#include <string.h>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <set>#include <stack>#include <string>#include <map>#include <time.h>#define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define EPS 1e-6#define FOR0(i,x) for(i=0;i<x;i++)#define FOR1(i,x) for(i=1;i<=x;i++)#define FOR(i,a,b) for(i=a;i<=b;i++)#define FORL0(i,a) for(i=a;i>=0;i--)#define FORL1(i,a) for(i=a;i>=1;i--)#define FORL(i,a,b)for(i=a;i>=b;i--)#define rush() int CC;for(scanf("%d",&CC);CC--;)#define Rush(n) while(scanf("%d",&n)!=-1)using namespace std;void RD(int &x){scanf("%d",&x);}void RD(i64 &x){scanf("%lld",&x);}void RD(u64 &x){scanf("%I64u",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(i64 &x,i64 &y){scanf("%lld%lld",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(i64 &x,i64 &y,i64 &z){scanf("%lld%lld%lld",&x,&y,&z);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;}void PR(int x) {printf("%d\n",x);}void PR(int x,int y) {printf("%d %d\n",x,y);}void PR(i64 x) {printf("%lld\n",x);}void PR(i64 x,i64 y) {printf("%lld %lld\n",x,y);}void PR(u32 x) {printf("%u\n",x);}void PR(u64 x) {printf("%llu\n",x);}void PR(double x) {printf("%.2lf\n",x);}void PR(double x,double y) {printf("%.5lf %.5lf\n",x,y);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout<<x<<endl;}const int mod=10007;const i64 inf=((i64)1)<<40;const double dinf=1000000000000000000.0;const int INF=100000000;const int N=55;const int M=2005;int val[N],visit[N],n,m;int a[N],b[N];int c[N][N],d[N][N],num[N];int dp[N][M],f[N][M][N*2];void DFS(int x){ int i,j,k,t; if(!num[x]) { b[x]=min(b[x],m/a[x]); for(i=0;i<=b[x];i++)for(j=0;j<=i;j++) { f[x][i*a[x]][j]=val[x]*(i-j); } return; } b[x]=INF; for(i=1;i<=num[x];i++) { k=c[x][i]; DFS(k); b[x]=min(b[x],b[k]/d[x][i]); a[x]+=d[x][i]*a[k]; } b[x]=min(b[x],m/a[x]); for(t=0;t<=b[x];t++) { clr(dp,-0x3f3f3f3f); dp[0][0]=0; for(i=1;i<=num[x];i++) for(j=0;j<=m;j++) { for(k=0;k<=j;k++) { dp[i][j]=max(dp[i][j],dp[i-1][j-k]+f[c[x][i]][k][t*d[x][i]]); } } for(j=0;j<=m;j++) for(k=0;k<=t;k++) { f[x][j][k]=max(f[x][j][k],dp[num[x]][j]+val[x]*(t-k)); } }}int main(){ RD(n,m); int i,j; char s[5]; FOR1(i,n) { RD(val[i]); RD(s); if(s[0]=='A') { RD(num[i]); FOR1(j,num[i]) { RD(c[i][j],d[i][j]); visit[c[i][j]]=1; } } else { RD(a[i],b[i]); } } FOR1(i,n) if(!visit[i]) break; int root=i; clr(f,-0x3f3f3f3f); DFS(root); int ans=0; for(i=0;i<=m;i++) for(j=0;j<=b[root];j++) { ans=max(ans,f[root][i][j]); } PR(ans);}