【BZOJ1563】诗人小G（决策单调性DP）

题意：给定N，L，P，求f[N]

sum[i]递增，L<=3e6，P<=10

思路：四边形不等式的证明见https://www.byvoid.com/zhs/blog/noi-2009-poet

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef long double ld;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 //typedef pair<ll,ll>P;
 #define N  300010
 #define M  200010
 #define INF 1e18
 #define fi first
 #define se second
 #define MP make_pair
 #define pb push_back
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1
 
 const //ll MOD=1e9+7,inv2=(MOD+1)/2;
       double eps=1e-;
       //int INF=1e9;
       int dx[]={-,,,};
       int dy[]={,,-,};
 
 struct node
 {
     int l,r,id;
 }q[N];
 
 int n,L,P;
 ld dp[N],s[N];
 char ch[];
 
 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }
 
 ld pw(ld x)
 {
     if(x<) x=-x;
     ld ans=;
     rep(i,,P) ans*=x;
     return ans;
 }
 
 ld calc(int j,int i)
 {
     return dp[j]+pw(s[i]-s[j]-L+i-j-);
 }
 
 void solve()
 {
     n=read(),L=read(),P=read();
     s[]=;
     rep(i,,n)
     {
         scanf("%s",ch+);
         int x=strlen(ch+);
         s[i]=s[i-]+x;
     }
     int h=,t=,l,r,last;
     q[]={,n,};
     rep(i,,n)
     {
         while(i>q[h].r) h++;
         dp[i]=calc(q[h].id,i);
         if(calc(i,n)>calc(q[t].id,n)) continue;
         while(i<q[t].l&&calc(i,q[t].l)<calc(q[t].id,q[t].l)) t--;
         l=max(q[t].l,i+);
         r=q[t].r;
         last=min(n,q[t].r+);
         while(l<=r)
         {
             int mid=(l+r)>>;
             if(calc(i,mid)<calc(q[t].id,mid)){last=mid; r=mid-;}
              else l=mid+;
         }
         q[t].r=last-;
         q[++t]={last,n,i};
     }
     if(dp[n]>INF) printf("Too hard to arrange\n");
      else printf("%lld\n",(ll)dp[n]);
     printf("--------------------\n");
 }
 
 int main()
 {
     int cas=read();
     while(cas--) solve();
     return ;
 }