description

solution

題目的意思是板凳不可重疊，資料不能直接摘蘋果。對蘋果排序，對凳子按r從小到大排序。設f[i][j]表示前i個凳子，選了j個，最後一個選了i的最大貢獻，列舉由那個f[k][j-1]轉移過來，能貢獻多少就在蘋果序上二分再max一下i左端點k右端點+1即可。

code

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LF double
#define LL long long
#define ULL unsigned int
 

#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=1e6+3,mm=200+3,inf=1e9+7;
int n,m,h,K,a[mn],f[mm][mm];
struct rec{
    int l,r;
};
rec b[mm];
bool cmp(rec x,rec y){
    return x.r<y.r;
}
int
 
 main(){
    freopen("apple.in","r",stdin);
    freopen("apple.out","w",stdout);
    scanf("%d%d%d%d",&n,&m,&h,&K);
    fo(i,1,n)scanf("%d",&a[i]);
    a[n+1]=inf;
    fo(i,1,m)scanf("%d%d",&b[i].l,&b[i].r);
    sort(a+1,a+n+1);
    fo(i,1,m){
        b[i].l=lower_bound(a+1
 
,a+n+2,h+b[i].l)-a;
        b[i].r=lower_bound(a+1,a+n+2,h+b[i].r+1)-a-1;
    }
    sort(b+1,b+m+1,cmp);
    fo(i,1,m)fo(j,1,K){
        f[i][j]=f[i-1][j];
        fo(k,0,i-1)
            f[i][j]=max(f[i][j],f[k][j-1]+max(b[i].r-max(b[i].l,b[k].r+1)+1,0));
    }
    printf("%d",f[m][K]);
    return 0;
}