斜率优化

套路：把转移方程推到只有关$i,j$,与$i,j$的乘积时(由$j$转移到$i$),把只与$i$有关的量放到一起作截距$B$,只与$j$相关的量作已知点的纵坐标$Y$,$i,j$的乘积中常量与$i$相关量的乘积作斜率$K$,$j$相关量作已知点的横坐标$X$，再进行线性规划

1.P2365任务安排$(O(n^2)$可过题）
一批任务的准备时间会影响从这批任务开始的所有任务的结束时间

\[f[i]=min_{0\le j<i}\{f[j]+SumT[i]*(SumC[i]-SumC[j])+s*(SumC[n]-SumC[j])\} \]

[BZOJ2726][SDOI2012]任务安排（斜率优化+$cdq$分治）~~cdq分治我们不熟~~

2.[APIO2010]特别行动队

\[f_i=max_{0\le j<i}\{f_j+a(s_i-s_j)^2+b(s_i-s_j)+c\} \]

变形一下

\[f_i=f_j+as_i^2-2as_is_j+as_j^2+bs_i-bs_j+c \]

\[f_j-as_j^2+bs_j=-2as_is_j+f_i+as_i^2+bs_i \]

令$$X_j=s_j$$ $$Y_j=f_j-as_j^2+bs_j$$

\[K_i=-2as_i$$ $$B_i=f_i+as_i^2+bs_i \]

$\because a < 0$ ，$\therefore K_i < 0$, 且$K$递增，要使$B$最大，维护上凸包

double eps=1e-7;
int dcmp(double x){if(fabs(x) <= eps) return 0; return x > 0 ? 1 : -1;}
double Y(int x){return f[x]+a*s[x]*s[x]-b*s[x];}
double X(int x){return s[x];}
double slope(int i, int j){
		double x1=X(i), y1=Y(i), x2=X(j), y2=Y(j);
		return (y1-y2)/(x1-x2);
}
int q[Maxn], l, r;
void solve(){
		n=read(); a=read(), b=read(), c=read(); for(int i=1; i <= n; i++) s[i]=s[i-1]+read();
		l=r=1, q[1]=0;
		for(int i=1; i <= n; i++){
			while(l < r && dcmp(slope(q[l], q[l+1]) - 1.0*a*2*s[i]) > 0) l++;
			f[i]=f[q[l]]+a*(s[i]-s[q[l]])*(s[i]-s[q[l]])+b*(s[i]-s[q[l]])+c;
			while(l < r && dcmp(slope(q[r-1], q[r])-slope(q[r], i) <= 0)) r--;
			q[++r]=i;
		}
		printf("%.0lf", f[n]);
}

3.[HNOI2008]玩具装箱TOY
令$S_i=\sum_{j=1}^iC_j$,$P_i=S_i+i$
则$$f_i=min_{0\le j<i}{f_j+(P_i-P_j-1-L)^2}$$

\[f_i=f_j+P_i^2-2P_iP_j+P_j^2-2(L+1)(P_i-P_j)+(L+1)^2 \]

\[f_j+P_j^2=2(P_i-L-1)P_j+f_i-P_i^2+2(L+1)P_i+(L+1)^2 \]

令$$X_j=P_j$$ $$Y_j=f_j+P_j^2$$

\[K_i=2(P_i-L-1)$$ $$B_i=f_i-P_i^2+2(L+1)P_i+(L+1)^2 \]

$\because P_i > 0,\therefore K_i$递增，要使$B_i$最小，维护下凸包

#define a(i) (sum[i]+i)
#define b(i) (a(i)+L+1)
#define Y(i) (dp[i]+b(i)*b(i))
#define X(i) b(i)
#define slope(i, j) ((Y(i)-Y(j))/(X(i)-X(j)))
int q[N], head, tail;
void solve(){
		n=read(), L=read(); head=tail=1;
		for(int i=1; i <= n; i++){
			sum[i]=sum[i-1]+read();
			while(head < tail && slope(q[head], q[head+1]) < 2*a(i)) head++;
			dp[i]=dp[q[head]]+(a(i)-b(q[head]))*(a(i)-b(q[head]));
			while(head < tail && slope(i, q[tail-1]) < slope(q[tail-1], q[tail])) tail--;
			q[++tail]=i;
		}
		printf("%lld\n", dp[n]);
}

4.[SDOI2016]征途
令$S_i$为每天的路程长度，$sum$为总路程，先推$vm^2$

\[\Delta S=\frac{sum}{m} \]

\[vm^2=m\sum_{i=1}^m(S_i-\Delta S)^2 \]

\[=m\sum_{i=1}^mS_i^2-2\Delta SS_i+\Delta S^2 \]

\[=m\sum_{i=1}^mS_i^2-sum^2 \]

所以令$f_{(i, j)}$为前$i$段路分为$j$天走的最小$\sum_{i=1}^jS_i$

\[f_{(i, j)}=min_{j-1\le k<i}\{f_{(k, j-1)}+(Sum_{i}-Sum_k)^2\} \]

不管$j$,

\[f_i=f_k+Sum_i^2-2Sum_iSum_k+Sum_k^2 \]

\[f_k+Sum_k^2=f_i-Sum_i^2+2Sum_iSum_k \]

令$$X_k=Sum_k$$ $$Y_k=f_k+Sum_k^2$$

\[K_i=2Sum_i \]

\[B_i=f_i-Sum_i^2 \]

令$B_i$最小，$\because K_i$递增，$\therefore$维护下凸包

int q[Maxn], l, r;
ll *f=f1, *g=g1;
double X(int i) {return S[i];}
double Y(int i) {return g[i]+S[i]*S[i];}
double slope(int i, int j){
	double x1=X(i), y1=Y(i), x2=X(j), y2=Y(j);
	return (y1-y2)/(x1-x2);
}
void solve(){
		n=read(), m=read(); for(int i=1; i <= n; i++) S[i]=S[i-1]+read(), f[i]=S[i]*S[i];
		for(int i=1; i < m; i++){
			l=r=1; q[1]=i; swap(f, g);
			for(int j=i+1; j <= n; j++){
				while(l < r && slope(q[l], q[l+1]) <= 2.0*S[j]) l++;
				f[j]=g[q[l]]+(S[j]-S[q[l]])*(S[j]-S[q[l]]);
				while(l < r && slope(q[r], q[r-1]) >= slope(q[r], j)) r--;
				q[++r]=j;
			}
		}
		printf("%lld", f[n]*m-S[n]*S[n]);
}

5.[CF311B]Cats Transport
令 $$d_i=\sum_{j=1}^iD_j, A_i=T_i-d_i$$
则$A_i$为接到猫$i$的最早出发时刻，若出发时刻为$t$,则猫等待的时间为$t-A_i$，把猫按$A$从小到大排序，令$S_i=\sum_{j=1}^iA_i$
，则有

每个$feeder$接到的为连续一段的猫最优

证明略~~(懒)~~

\[f_{i,j}=min_{0\le k\le i}\{f_{k, j-1}+A_i(i-k)-(S_i-S_k)\} \]

不管$j$

\[f_i=f'_k+iA_i-kA_i-S_i+S_k \]

\[f_i-iA_i+S_i+kA_i=f'_k+S_k \]

令$$X_k=k$$ $$Y_k=f'_k+S_k$$

\[K_i=A_i$$ $$B_i=f_i-iA_i+S_i \]

要令$B$最小，$\because K$递增，$\therefore$维护下凸包

ll *f=f1, *g=g1;
int q[Maxn], l, r;
double X(int i){return i;}
double Y(int i){return g[i]+S[i];}
double slope(int i, int j){
    	double x1=X(i), x2=X(j), y1=Y(i), y2=Y(j);
    	return (y2-y1)/(x2-x1);
}
void solve(){
    	n=read(), m=read(), p=read();
    	for(int i=2; i <= n; i++) D[i]=D[i-1]+read();
    	for(int i=1, x, t; i <= m; i++) x=read(), t=read(), A[i]=t-D[x];
    	sort(A+1, A+1+m); for(int i=1; i <= m; i++) S[i]=S[i-1]+A[i];
    	for(int i=1; i <= m; i++) f[i]=A[i]*i-S[i];
    	for(int i=1; i < p; i++){
        	q[1]=0; l=r=1; swap(f, g);
        	for(int j=1; j <= m; j++){
            	while(l < r && slope(q[l], q[l+1]) <= A[j]) l++;
            	f[j]=g[q[l]]+(j-q[l])*A[j]+S[q[l]]-S[j];
            	while(l < r && slope(q[r], q[r-1]) >= slope(q[r], j)) r--;
            	q[++r]=j;
        	}
   		}
    	printf("%lld", f[m]);
}

6.[POJ3709]K-Anonymous Sequence
- 题意:求通过减小序列中元素的方式把给定序列$a[]$变为$K-Anonymous$序列的最小代价
- 定义:
1. $K-Anonymous$：对于$\forall s \in a[]$,至少存在其它$k-1$个元素与它相等
2. 代价:把元素$s$变为$s'$的代价为$s-s'$

\[f_i=min_{0\le j\le i-k}\{f_j+S_i-S_j-(i-j)a_{j+1}\} \]

\[f_j-S_j+ja_{j+1}=f_i-S_i+ia_{j+1} \]

令$$X_j=a_{j+1}$$ $$Y_j=f_j-S_j+ja_{j+1}$$

\[K_i=i$$ $$B_i=f_i-S_i \]

要使$B$最小,$\because K$递增，$\therefore$维护下凸包(我不知道为什么用$double$求$slope$会被卡)

ll Y(int i){return f[i]-S[i]+a[i]*i;}
int q[Maxn], l, r;
void solve(){
		int T=read();
		while(T--){
			n=read(), k=read(); for(int i=0; i < n; i++) a[i]=read();
			for(int i=1; i <= n; i++) S[i]=S[i-1]+a[i-1];
			l=r=1, q[1]=0;
			for(int i=1; i <= n; i++){
				while(l < r && (a[q[l+1]]-a[q[l]])*i >= Y(q[l+1])-Y(q[l])) l++;
				f[i]=f[q[l]]+S[i]-S[q[l]]+a[q[l]]*(q[l]-i);
				if(i-k+1 >= k) {
					while(l < r && (a[q[r-1]]-a[q[r]])*(Y(q[r])-Y(i-k+1)) <= (a[q[r]]-a[i-k+1])*(Y(q[r-1])-Y(q[r]))) r--; 
					q[++r]=i-k+1; 
				}
			}
			printf("%lld\n", f[n]);
		}
}

posted @ 2018-07-08 21:43 zerolt 阅读(197) 评论(0) 编辑收藏举报

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斜率优化

每个\(feeder\)接到的为连续一段的猫最优

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