HDU3401【Trade】

Trade

topic

Trade

---

Parsing

Establish DP first, and then perform monotonic queue optimization.
Be careful! ! ! The initial value must be set! ! !
Lessons of blood and tears

code:

#include<deque>
#include<cstdio>
#include<cstring>
using namespace std;
int max(int x,int y){
    
    return x>y?x:y;}
int T,n,p,w,a[2010],b[2010],as[2010],bs[2010],dp[2010][2010],ans,k,l;
bool idigit(char x){
    
    return (x<'0'|x>'9')?0:1;}
inline int read()
{
    
    
	int num=0,f=1;
	char c=0;
	while(!idigit(c=getchar())){
    
    if(c=='-')f=-1;}
	while(idigit(c))num=(num<<1)+(num<<3)+(c&15),c=getchar();
	return num*f;
}
inline void write(int x)
{
    
    
	int F[20];
	int tmp=x>0?x:-x;
	if(x<0)putchar('-');
	int cnt=0;
	while(tmp>0){
    
    F[cnt++]=tmp%10+'0';tmp/=10;}
	while(cnt>0)putchar(F[--cnt]);
	if(x==0)putchar('0');
	putchar('\n');
}
struct f
{
    
    
	int x,y;
}o2;
deque <f> c;
int main()
{
    
    
	T=read();
	while(T--)
	{
    
    
		n=read(),p=read(),w=read();
		++w,ans=0;
		for(int i=1;i<=n;i++)a[i]=read(),b[i]=read(),as[i]=read(),bs[i]=read();
		memset(dp,0xcf,sizeof(dp));
		for(int i=1;i<=w;i++)for(int j=0;j<=as[i];j++)dp[i][j]=-j*a[i];//置初值
		for(int i=1;i<=n;i++)
		{
    
    
			for(int j=0;j<=p;j++)dp[i][j]=max(dp[i][j],dp[i-1][j]);
			if(i<=w)continue;
			k=i-w;
			c.clear();
			for(int j=0;j<=p;j++)
			{
    
    
				l=dp[k][j]+j*a[i];
				while((!c.empty())&&c.back().x<l)c.pop_back();
				o2.x=l,o2.y=j;
				c.push_back(o2);
				while((!c.empty())&&c.front().y+as[i]<j)c.pop_front();
				dp[i][j]=max(dp[i][j],c.front().x-j*a[i]);
			}
			c.clear();
			for(int j=p;j>=0;j--)
			{
    
    
				l=dp[k][j]+j*b[i];
				while((!c.empty())&&c.back().x<l)c.pop_back();
				o2.x=l,o2.y=j;
				c.push_back(o2);
				while((!c.empty())&&c.front().y-bs[i]>j)c.pop_front();
				dp[i][j]=max(dp[i][j],c.front().x-j*b[i]);
			}//单调队列优化DP
		}
		for(int i=0;i<=p;i++)if(ans<dp[n][i])ans=dp[n][i];//可能取最值
		write(ans);
	}
	return 0;
}