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
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned int uint; 5 typedef unsigned long long ull; 6 typedef long double ld; 7 typedef pair<int,int> PII; 8 typedef pair<ll,ll> Pll; 9 typedef vector<int> VI; 10 typedef vector<PII> VII; 11 //typedef pair<ll,ll>P; 12 #define N 300010 13 #define M 200010 14 #define INF 1e18 15 #define fi first 16 #define se second 17 #define MP make_pair 18 #define pb push_back 19 #define pi acos(-1) 20 #define mem(a,b) memset(a,b,sizeof(a)) 21 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++) 22 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--) 23 #define lowbit(x) x&(-x) 24 #define Rand (rand()*(1<<16)+rand()) 25 #define id(x) ((x)<=B?(x):m-n/(x)+1) 26 #define ls p<<1 27 #define rs p<<1|1 28 29 const //ll MOD=1e9+7,inv2=(MOD+1)/2; 30 double eps=1e-6; 31 //int INF=1e9; 32 int dx[4]={-1,1,0,0}; 33 int dy[4]={0,0,-1,1}; 34 35 struct node 36 { 37 int l,r,id; 38 }q[N]; 39 40 int n,L,P; 41 ld dp[N],s[N]; 42 char ch[100]; 43 44 int read() 45 { 46 int v=0,f=1; 47 char c=getchar(); 48 while(c<48||57<c) {if(c=='-') f=-1; c=getchar();} 49 while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar(); 50 return v*f; 51 } 52 53 ld pw(ld x) 54 { 55 if(x<0) x=-x; 56 ld ans=1; 57 rep(i,1,P) ans*=x; 58 return ans; 59 } 60 61 ld calc(int j,int i) 62 { 63 return dp[j]+pw(s[i]-s[j]-L+i-j-1); 64 } 65 66 void solve() 67 { 68 n=read(),L=read(),P=read(); 69 s[0]=0; 70 rep(i,1,n) 71 { 72 scanf("%s",ch+1); 73 int x=strlen(ch+1); 74 s[i]=s[i-1]+x; 75 } 76 int h=0,t=0,l,r,last; 77 q[0]={1,n,0}; 78 rep(i,1,n) 79 { 80 while(i>q[h].r) h++; 81 dp[i]=calc(q[h].id,i); 82 if(calc(i,n)>calc(q[t].id,n)) continue; 83 while(i<q[t].l&&calc(i,q[t].l)<calc(q[t].id,q[t].l)) t--; 84 l=max(q[t].l,i+1); 85 r=q[t].r; 86 last=min(n,q[t].r+1); 87 while(l<=r) 88 { 89 int mid=(l+r)>>1; 90 if(calc(i,mid)<calc(q[t].id,mid)){last=mid; r=mid-1;} 91 else l=mid+1; 92 } 93 q[t].r=last-1; 94 q[++t]={last,n,i}; 95 } 96 if(dp[n]>INF) printf("Too hard to arrange "); 97 else printf("%lld ",(ll)dp[n]); 98 printf("-------------------- "); 99 } 100 101 int main() 102 { 103 int cas=read(); 104 while(cas--) solve(); 105 return 0; 106 }