【jzoj5350】【NOIP2017提高A组模拟9.7】【陶陶摘苹果】【动态规划】

description

solution

题目的意思是板凳不可重叠，数据不能直接摘苹果。对苹果排序，对凳子按r从小到大排序。设f[i][j]表示前i个凳子，选了j个，最后一个选了i的最大贡献，枚举由那个f[k][j-1]转移过来，能贡献多少就在苹果序上二分再max一下i左端点k右端点+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;}