Description

有n个点，每个点有时间及价值，斜率相同的点存在冲突，要先取近的才能取远的，求T的时间内能得到的最大价值

Code

#include<bits/stdc++.h>
using namespace std;
const int N=202;
struct node{
	int x,y,len,w,v;
}s[N];
struct kk{
	int w,v;
}tmp;
bool cmp(node a,node b){return a.len<b.len;}
vector<kk>se[N];
bool mk[N];
int dp[N*N],n,i,j,T,t,k,w,v;
int main(){
	scanf("%d%d",&n,&T);
	for (i=0;i<n;i++){
		scanf("%d%d%d%d",&s[i].x,&s[i].y,&s[i].w,&s[i].v);
		s[i].len=s[i].x*s[i].x+s[i].y*s[i].y;
	}
	sort(s,s+n,cmp);
	for (i=0;i<n;i++){
		if (mk[i]) continue;
		se[k].clear();
		tmp.w=s[i].w;
		tmp.v=s[i].v;
		se[k].push_back(tmp);
		mk[i]=1;
		for (j=i+1;j<n;j++){
			if (mk[j])continue;
			if (s[i].y*s[j].x==s[i].x*s[j].y){
				tmp.w+=s[j].w;
				tmp.v+=s[j].v;
				se[k].push_back(tmp);
				mk[j]=1;
			}
		}
		k++;
	}
	for (i=0;i<k;i++)
		for (j=T;j>=0;j--)
			for (t=0;t<se[i].size() && se[i][t].w<=j;t++) dp[j]=max(dp[j],dp[j-se[i][t].w]+se[i][t].v);
	printf("%d",dp[T]);
}