#斜率優化dp# [poj 1180] Batch Scheduling

題目

http://poj.org/problem?id=1180

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解題思路

$n^2$的$dp:$ $f[i]=min(f[j]+sumT[i]*(sumC[i]-sumC[j])+S*(sumC[N]-sumC[j]))$
$f[j]=(S+sumT[i])*sumC[j]+f[i]-sumT[i]*sumC[i]-S*sumC[n]$
每個待求解的狀態$f[i]$都對應着一條直線的截距，直線的斜率是$S+sumT[i]$，截距未知。當截距最小化，$f[i]$也取到最小值。
有三個決策點$j1,j2,j3(j1<j2<j3)$
$j2$有可能成爲最優決策，當且僅當$\frac{f[j2]-f[j1]}{sumC[j2]-sumC[j1]}<\frac{f[j3]-f[j2]}{sumC[j3]-sumC[j2]}$
用單調隊列維護$“下凸殼”$

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代碼

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std; 
int n,s,t,c,sumt[10001],sumc[10001],f[10001],q[10001];
int main(){
	scanf("%d%d",&n,&s); 
	for(int i=1;i<=n;i++) scanf("%d%d",&t,&c),sumt[i]=sumt[i-1]+t,sumc[i]=sumc[i-1]+c; 
	memset(f,0x3f,sizeof(f)); 
	int l=1,r=1; q[1]=0; f[0]=0;
	for(int i=1;i<=n;i++){
		while(l<r&&(f[q[l+1]]-f[q[l]])<=(s+sumt[i])*(sumc[q[l+1]]-sumc[q[l]])) l++; 
		f[i]=f[q[l]]-(s+sumt[i])*sumc[q[l]]+sumt[i]*sumc[i]+s*sumc[n]; 
		while(l<r&&(f[q[r]]-f[q[r-1]])*(sumc[i]-sumc[q[r]])>=(f[i]-f[q[r]])*(sumc[q[r]]-sumc[q[r-1]])) r--; 
		q[++r]=i; 
	}
	printf("%d",f[n]); 
}