Machine Works UVALive - 5133

Machine Works UVALive - 5133

超哥线段树优化Dp

题解：

裸方程很好想：

$$f_i=\max\{f_j+G_j*(D_i-D_j-1\}-P_i+R_i$$

当前面的max小于Pi的时候$f_i=-\inf$.

时间复杂度：$O(n^2)$.

其实就是维护一次函数的最大值。

直接超哥线段树即可。

正解好像是单调队列 ？

Code：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#define D(x) cout<<#x<<" = "<<x<<"  "
#define E cout<<endl
using namespace std;
typedef long long ll;
typedef long double ld;
const ll INF = 1e18;
const int N = 100005;
const int mxn = 1e9;

ll C,D,f[N];
int n,sz,root,lch[N*30],rch[N*30];
ll K[N*30],B[N*30];
struct Machine{
    ll d,p,r,g;
} m[N];
bool cmp(const Machine &a, const Machine &b){
    return a.d < b.d;
}

inline ll getB(ll k,ll x,ll y){ return y-k*x; }

inline bool better(ll k1,ll b1,ll k2,ll b2,ll x){ return k1*x+b1 > k2*x+b2; }

void insert(int &x,int l,int r,ll k,ll b){
    if(!x) x=++sz;
    if(l==r){
        if(better(k,b,K[x],B[x],l)){ K[x]=k; B[x]=b; }
        return;
    }
    else{
        int mid=(l+r)>>1;
        if(k > K[x]){
            if(better(k,b,K[x],B[x],mid)){
                swap(K[x],k); swap(B[x],b);
                insert(lch[x],l,mid,k,b);
            }
            else insert(rch[x],mid+1,r,k,b);
        }
        else{
            if(better(k,b,K[x],B[x],mid)){
                swap(K[x],k); swap(B[x],b);
                insert(rch[x],mid+1,r,k,b);
            }
            else insert(lch[x],l,mid,k,b);
        }
    }
}

ll query(int x,int l,int r,int p){
    if(!x) return 0.0;
    ll d=K[x]*p+B[x];
    if(l==r) return d;
    else{
        int mid=(l+r)>>1;
        if(p<=mid) return max(d,query(lch[x],l,mid,p));
        else return max(d,query(rch[x],mid+1,r,p));
    }
}

int main(){
    freopen("a.in","r",stdin); 
    freopen("a.out","w",stdout);
    int cas=0;
    while(scanf("%d%lld%lld",&n,&C,&D)!=EOF && (n||C||D)){
        for(int i=1;i<=n;i++){
            scanf("%lld%lld%lld%lld",&m[i].d,&m[i].p,&m[i].r,&m[i].g);
        }
        sort(m+1,m+1+n,cmp);
//      for(int i=1;i<=n;i++){
//          printf("d = %d  p = %d  r = %d  g = %d\n",m[i].d,m[i].p,m[i].r,m[i].g);
//      }
        insert(root,1,mxn,0,C);
        int last=0;
        for(int i=1;i<=n;i++){
            if(i>1 && m[i-1].d!=m[i].d){
                for(int j=last+1;j<=i-1;j++){
                    if(f[j]!=-1) insert(root,1,mxn,m[j].g,getB(m[j].g,m[j].d,f[j]));
                }
                last=i-1;
            }
            f[i]=query(root,1,mxn,m[i].d-1);
//          D(f[i]);
            if(f[i]>=m[i].p) f[i]+=m[i].r-m[i].p;
            else f[i]=-1;
//          D(f[i]); E;
//          cout<<f[i]<<endl;
        }
        ll ans=C;
        for(int i=1;i<=n;i++){
            if(f[i]!=-1) ans=max(ans, f[i]+m[i].g*(D-m[i].d));
        }
        printf("Case %d: %lld\n",++cas,ans);
    }
}