华为oj 购物单

这两天断断续续敲完这个（放假的时候比较懒），一次成功有点小激动(●'◡'●) 不过貌似从第一次打开开始计时。。。。。

这道题目很像01背包，我将附件与它们的主件绑定（就是link起来）然后套用动态规划

ok，解决

#include<iostream>
#include<vector>
using namespace std;

class node {//代表主件的类
public:
int value;
int p;//重要度
int n;//附件个数
int q;//附件指向的主件编号
node *link;
node():value(0),p(0),n(0),q(0),link(nullptr){}
};

int judge(vector<node*>& vc,int N,int m);
int max(int, int);

int main()
{
int N, m;
vector<node*> vc;
cin >> N >> m;
for (int i = 0;i < m;i++) {
int vi, pi, qi;
cin >> vi >> pi >> qi;
node *np = new node();
np->value = vi;
np->p = pi;
np->q = qi;
vc.push_back(np);
}
for (vector<node*>::iterator i = vc.begin();i != vc.end();i++) {
int tmp = (*i)->q;
if (tmp > 0) {
if(vc[tmp-1]->link==nullptr)
vc[tmp - 1]->link = *i;
else
vc[tmp - 1]->link->link = *i;
vc[tmp - 1]->n++;
}

}
for (vector<node*>::iterator i = vc.begin();i != vc.end();) {
int tmp = (*i)->q;
if (tmp > 0) {
i=vc.erase(i);
}
else {
++i;
}

}
cout<<judge(vc, N, m)<<endl;
return 0;
}
int judge(vector<node*>& vc, int N, int m) {
//任务 返回经过计算的结果
//建立一个int[3200][60]数组 因为N为10的倍数
int tb[3200][60] = {0};
vector<node*>::iterator ib = vc.begin();
for (int i = 1;i <= N / 10;i++) {
if ((*ib)->n == 0) {
if ((*ib)->value <= i*10) {
tb[i][1] = ((*ib)->value*(*ib)->p);
}
}
if ((*ib)->n == 1) {
if (((*ib)->value+(*ib)->link->value) <= i*10) {
tb[i][1] = ((*ib)->value*(*ib)->p)+ ((*ib)->link->value*(*ib)->link->p);
}
else {
tb[i][1] = ((*ib)->value*(*ib)->p);
}
}
if ((*ib)->n == 2) {
if (((*ib)->value + (*ib)->link->value+ (*ib)->link->link->value) <= i*10) {
tb[i][1] = ((*ib)->value*(*ib)->p) + ((*ib)->link->value*(*ib)->link->p)+ ((*ib)->link->link->value*(*ib)->link->link->p);
}
else if (((*ib)->value + (*ib)->link->value) <= i*10) {
tb[i][1] = ((*ib)->value*(*ib)->p) + ((*ib)->link->value*(*ib)->link->p);
}
else {
tb[i][1] = ((*ib)->value*(*ib)->p);
}
}
}
++ib;//使迭代器指向下一个元素
int j=1;
for (;ib != vc.end();ib++,j++) {
for (int i = 1;i <= N / 10;i++) {

if ((*ib)->n == 0) {//没有附件
int tmp1 = (*ib)->value;
int tmp1_ = (*ib)->p;
if (tmp1 > i * 10) {
tb[i][j] = tb[i][j - 1];
}
else {
tb[i][j] = max(tb[i][j - 1], tb[i - tmp1 / 10][j - 1] + tmp1*tmp1_);
}
}

if ((*ib)->n == 1) {//有一个附件
int tmp1 = (*ib)->value;
int tmp1_ = (*ib)->p;
int tmp2 = (*ib)->link->value;
int tmp2_ = (*ib)->link->p;

if (tmp1 > i * 10) {
tb[i][j] = tb[i][j - 1];
}
else if ((tmp1 + tmp2) > i * 10) {
tb[i][j] = max(tb[i][j - 1], tb[i - tmp1 / 10][j - 1] + tmp1*tmp1_);
}
else {
int m1= max(tb[i][j - 1], tb[i - tmp1 / 10][j - 1] + tmp1*tmp1_);
tb[i][j] = max(m1, tb[i - (tmp1+tmp2) / 10][j - 1] + tmp1*tmp1_+tmp2*tmp2_);
}
}

if ((*ib)->n == 1) {//有两个附件
int tmp1 = (*ib)->value;
int tmp1_ = (*ib)->p;
int tmp2 = (*ib)->link->value;
int tmp2_ = (*ib)->link->p;
int tmp3= (*ib)->link->link->value;
int tmp3_ = (*ib)->link->link->p;
if (tmp1 > i * 10) {
tb[i][j] = tb[i][j - 1];
}
else if ((tmp1 + tmp2) > i * 10) {
tb[i][j] = max(tb[i][j - 1], tb[i - tmp1 / 10][j - 1] + tmp1*tmp1_);
}
else if((tmp1+tmp2+tmp3)>i*10){
int m1 = max(tb[i][j - 1], tb[i - tmp1 / 10][j - 1] + tmp1*tmp1_);
tb[i][j] = max(m1, tb[i - (tmp1 + tmp2) / 10][j - 1] + tmp1*tmp1_ + tmp2*tmp2_);
}
else {
int m1 = max(tb[i][j - 1], tb[i - tmp1 / 10][j - 1] + tmp1*tmp1_);
int m2= max(m1, tb[i - (tmp1 + tmp2) / 10][j - 1] + tmp1*tmp1_ + tmp2*tmp2_);
tb[i][j] = max(m2, tb[i - (tmp1 + tmp2 + tmp3) / 10][j - 1] + tmp1*tmp1_ + tmp2*tmp2_ + tmp3*tmp3_);
}
}

}
}
return tb[N/10][j-1];

}

int max(int a, int b) {
if (a > b)
return a;
else
return b;
}