【动态规划】[luoguP1417]烹调方案

题目

我们要先推一个贪心
考虑相邻的两个物品xy 已经耗费p的时间
先做x
a[x]−(p+c[x])∗b[x]+a[y]−(p+c[x]+c[y])∗by ————①
先做y
a[y]−(p+c[y])∗b[y]+a[x]−(p+c[y]+c[x])∗bx ————②
所以①＞②的条件是

c[x]∗b[y]<c[y]∗b[x]

所以sort一下 然后就是01背包
关键这个题必须都得用long long！

代码如下

#include<iostream>#include<cstdio>#include<cctype>#include<algorithm>    using namespace std;    #define in = read();    typedef long long ll;    typedef unsigned int ui;    const ll size = 100000 + 1000;        struct point{   ll a , b , c;}data[size];            ll n , t;            ll ans;            ll f[size];inline ll read(){        ll num = 0 , f = 1;    char ch = getchar();        while(!isdigit(ch)){                if(ch == '-')   f = -1;                ch = getchar();        }        while(isdigit(ch)){                num = num*10 + ch - '0';                ch = getchar();        }        return num*f;}inline bool cmp(point x , point y){        return x.c*y.b < y.c*x.b;}int main(){        t in;   n in;        for(register ll i=1;i<=n;i++)   data[i].a in;        for(register ll i=1;i<=n;i++)   data[i].b in;        for(register ll i=1;i<=n;i++)   data[i].c in;        sort(data + 1 , data + n + 1 , cmp);        for(register ll i=1;i<=n;i++)                for(register ll j=t;j>=data[i].c;j--)                        f[j] = max(f[j] , f[j - data[i].c] + data[i].a - data[i].b*j);        for(register ll i=1;i<=t;i++)                ans = max(ans , f[i]);        printf("%d" , ans);}//COYG