动态规划与凹凸优化-优快云博客

见注释，注意一下由于要移项，所以x一定要是大减小！

任务安排 1 2 3

#include<bits/stdc++.h>
using namespace std;

#define read(x) scanf("%d",&x)
#define inf (1LL<<60)
#define maxn 5000 
#define maxS 50
#define maxw 100
#define ll long long

int n,S;
int t[maxn+5],c[maxn+5];
int sumt[maxn+5],sumc[maxn+5];

ll f[maxn+5];

int main() {
	
	read(n),read(S);
	for(int i=1;i<=n;i++) {
		read(t[i]),read(c[i]);
		sumt[i]=sumt[i-1]+t[i];
		sumc[i]=sumc[i-1]+c[i];
	}
	for(int i=1;i<=n;i++) {	//这一组在i结束 
		f[i]=inf;
		for(int j=0;j<i;j++) {	//上一组任务在j结束 
			f[i]=min(f[i],f[j]+sumt[i]*((ll)sumc[i]-sumc[j])+S*((ll)sumc[n]-sumc[j]));
		}
	}
	
	printf("%lld",f[n]);
		
	return 0;
}

#include<bits/stdc++.h>
using namespace std;

#define read(x) scanf("%d",&x)
#define inf (1LL<<60)
#define maxn int(3e5) 
#define maxS 512
#define maxw 512
#define ll long long

int n,S;
ll sumt[maxn+5],sumc[maxn+5]; 

ll f[maxn+5];
// f[i] = f[j] - (sumt[i]+S)*sumc[j] + sumt[i]*sumc[i] + S*sumc[n]
// f[j] = (S+sumt[i])*sumc[j] + f[i]-(sumt[i]+S)*sumc[i]

ll que[maxn+5];
int h,t; 

int main() {
	
	read(n),read(S);
	for(int i=1;i<=n;i++) {
		int t,c;
		read(t),read(c);
		sumt[i]=sumt[i-1]+t;
		sumc[i]=sumc[i-1]+c;
	}
	
	h=0,t=0;
	for(int i=1;i<=n;i++) {
		int k=S+sumt[i];	//当前点的斜率 
		while(h<t) {	//至少有两个数 
			int j1=que[h],j2=que[h+1];
			if(f[j2]-f[j1]<=(sumc[j2]-sumc[j1])*k) ++h;	// (f[j2]-f[j1]) / (sumc[j2]-sumc[j1]) <= k; 
			else break;
		}
		int j=que[h];
		f[i] = f[j] - (sumt[i]+S)*sumc[j] + sumt[i]*sumc[i] + S*sumc[n];
		while(h<t) {
			int j1=que[t-1],j2=que[t];
			if( (sumc[i]-sumc[j2])*(f[j2]-f[j1]) >= (sumc[j2]-sumc[j1])*(f[i]-f[j2]) ) --t;	// (f[j2]-f[j1]) / (sumc[j2]-sumc[j1]) >= (f[i]-f[j2]) / (sumc[i]-sumc[j2]); 
			else break;
		}
		que[++t]=i;
	}
	
	printf("%lld",f[n]);
	
		
	return 0;
}

#include<bits/stdc++.h>
using namespace std;

#define read(x) scanf("%d",&x)
#define inf (1LL<<60)
#define maxn int(3e5) 
#define maxS 512
#define maxw 512
#define ll long long

int n,S;
ll sumt[maxn+5],sumc[maxn+5]; 

ll f[maxn+5];
// f[i] = f[j] - (sumt[i]+S)*sumc[j] + sumt[i]*sumc[i] + S*sumc[n]
// f[j] = (S+sumt[i])*sumc[j] + f[i]-(sumt[i]+S)*sumc[i]

int que[maxn+5];
int h,t; 

bool check(int x,ll k) {
	return f[que[x+1]]-f[que[x]]<=(sumc[que[x+1]]-sumc[que[x]])*k;
}

int main() {
	
	read(n),read(S);
	for(int i=1;i<=n;i++) {
		int t,c;
		read(t),read(c);
		sumt[i]=sumt[i-1]+(ll)t;
		sumc[i]=sumc[i-1]+(ll)c;
	}
	
	h=0,t=0;
	for(int i=1;i<=n;i++) {
		ll k=S+sumt[i];	//当前点的斜率 
		
		int l=h,r=t;	//二分
		while(l<r) {
			int mid=(l+r)>>1;
			if(check(mid,k)) l=mid+1;
			else r=mid; 
		}
		
		int j=que[l]; 
		f[i] = f[j] - k*sumc[j] + sumt[i]*sumc[i] + S*sumc[n];
		while(h<t) {
			int j1=que[t-1],j2=que[t];
			if( (sumc[i]-sumc[j2])*((double)f[j2]-f[j1]) >= (sumc[j2]-sumc[j1])*((double)f[i]-f[j2]) ) --t;	// (f[j2]-f[j1]) / (sumc[j2]-sumc[j1]) >= (f[i]-f[j2]) / (sumc[i]-sumc[j2]); 
			else break;
		}
		que[++t]=i;
	}
	
	printf("%lld",f[n]);
	
		
	return 0;
}

运输小猫

#include<bits/stdc++.h>
using namespace std;

#define maxn 100005
#define maxp 105
#define inf (1LL<<60)
#define ll long long

int n,m,P;
ll D[maxn+5];	//sumD
ll d[maxn+5];	//sum{T[i]-sumD[H[i]]}

ll f[maxn+5][2];	//f[i] 前i只猫，p个运输员
//f[i][p] = f[j][p-1]+ (i-j)*(d[i]-d[i-1])-(d[i]-d[j])
//f[j][p-1]+d[j] = (d[i]-d[i-1])*j - (d[i]-d[i-1])*i + f[i][p]+d[i]

int que[maxn+5]; 
int h=0,t=0;

int main() {
	scanf("%d%d%d",&n,&m,&P);
	for(int i=2;i<=n;i++) {
		scanf("%lld",&D[i]);
		D[i]+=D[i-1]; 
	}
	for(int i=1;i<=m;i++) {
		int h,t;
		scanf("%d%d",&h,&t);
		d[i]=-D[h]+t;
	}
	
	sort(d+1,d+1+m);
	for(int i=1;i<=m;i++) d[i]+=d[i-1];
	
	for(int i=1;i<=m;i++) f[i][0]=inf;
	for(int p=1;p<=P;p++) {
		
		que[0]=que[1]=0;
		h=t=0;
		
		for(int i=1;i<=m;i++) {
			
			#define gety(x,y) (f[x][(y)&1]+d[x])
			
			ll k=d[i]-d[i-1];
			
			while(h<t) {
				int j1=que[h],j2=que[h+1];
				if(gety(j2,p-1) - gety(j1,p-1) <= k*(j2-j1)) ++h;
				else break;
			}
			
			int j=que[h];
			f[i][p&1] = f[j][(p-1)&1]+ (i-j)*k-(d[i]-d[j]);
			
			while(h<t) {
				int j1=que[t-1],j2=que[t];
				if( (gety(j2,p-1) - gety(j1,p-1))*(i-j2) >= (gety(i,p-1) - gety(j2,p-1))*(j2-j1) ) --t;
				else break;
			}
			que[++t]=i;
		}
	}
	
	printf("%lld",f[m][P&1]);
	
	return 0;
}

玩具装箱

#include<bits/stdc++.h>
using namespace std;

#define maxn ((int)5e4)

typedef long long ll;
typedef double db;

const db inf=1e30;

int n,L;
db sum[maxn+5]; 
db f[maxn+5];
//f[i] = f[j]+Square(sum[i]+i-sum[j]-j-1-L)
//f[j]+Square(sum[j]+j) = 2*(sum[i]+i-L-1)*(sum[j]+j) + f[i]-Square(sum[i]+i-L-1)

db Square(db x) {return x*x;}

int que[maxn+5];
int h,t;

int main() {
	
	scanf("%d%d",&n,&L);
	for(int i=1;i<=n;i++) {
		scanf("%lf",&sum[i]);
		sum[i]+=sum[i-1];
	}
	
	for(int i=1;i<=n;i++) {
		db k=2*(sum[i]+i-L-1); 
		
		#define x(i) (sum[i]+i)
		#define y(i) (f[i]+Square(sum[i]+i)) 
		
		while(h<t) {
			int j1=que[h],j2=que[h+1];
			if(y(j2)-y(j1)<=k*(x(j2)-x(j1))) ++h;
			else break;
		}
		
		int j=que[h];
		f[i] = f[j]+Square(sum[i]+i-sum[j]-j-1-L);
		
		while(h<t) {
			int j1=que[t-1],j2=que[t];
			if( (y(j2)-y(j1))*(x(i)-x(j2)) >= (y(i)-y(j2))*(x(j2)-x(j1)) ) --t;
			else break;
		}
		que[++t]=i;
	}
	
	printf("%.0lf",f[n]);
	
	return 0;
}