DP 斜率优化题目/决策单调性题目

斜率优化

冬令营讲到了dp斜率优化后瞎写了一些斜率优化，因为毕竟上次写是老早以前了，当时对这个并不是很理解，现在有一点数学基础，稍微好一点了。

瞎找的斜率优化题，可能有点太水了，我这种这么菜的人竟然都切得那么快。可能难度不够，因为有个y坐标不单调要套个平衡树的我是真不会，这个嘛以后再说了。

1.hdu3507 Print Article

给定非负数列，划成若干块，每块价值是左边那玩意儿。求总权值min.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,1:0;}
 5 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,1:0;}
 6 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 7 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 8 template<typename T>inline T read(T&x){
 9     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c==‘-‘)f=1;
10     while(isdigit(c))x=x*10+c-‘0‘,c=getchar();return f?x=-x:x;
11 }
12 const int N=500000+7;
13 ll f[N],C[N],m;
14 int q[N],l,r,n;
15 inline ll Getx(int i){return C[i];}
16 inline ll Gety(int i){return f[i]+C[i]*C[i];}
17
18 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
19     while(~scanf("%d%lld",&n,&m)){
20         q[l=r=1]=0,f[0]=0;
21         for(register int i=1;i<=n;++i){
22             read(C[i]),C[i]+=C[i-1];
23             while(l<r&&(Gety(q[l+1])-Gety(q[l]))<=2*C[i]*(Getx(q[l+1])-Getx(q[l])))++l;
24             f[i]=f[q[l]]+(C[i]-C[q[l]])*(C[i]-C[q[l]])+m;
25             while(l<r&&(Gety(q[r])-Gety(q[r-1]))*(Getx(i)-Getx(q[r]))>=(Gety(i)-Gety(q[r]))*(Getx(q[r])-Getx(q[r-1])))--r;
26             q[++r]=i;
27         }
28         printf("%lld\n",f[n]);
29     }
30     return 0;
31 }

2.P2365 任务安排

注意一个费用提前计算的思想，就是为了简化维度，并没有明显规定要画成多少块，但块数对f有影响，那可以提前计算对之后的费用加上去，这也应当是对的dp.详见李煜东书上的dp斜率优化。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N=5000+7;
 5 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,1:0;}
 6 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,1:0;}
 7 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 8 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 9 template<typename T>inline T read(T&x){
10     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c==‘-‘)f=1;
11     while(isdigit(c))x=x*10+c-‘0‘,c=getchar();return f?x=-x:x;
12 }
13 ll f[N],t[N],c[N],S;
14 int l,r,n,q[N];
15
16 int main(){//freopen("tmp.in","r",stdin);freopen("tmp.out","w",stdout);
17     read(n),read(S);
18     for(register int i=1;i<=n;++i)t[i]=read(t[i])+t[i-1],c[i]=read(c[i])+c[i-1];
19     f[l=r=1]=0;
20     for(register int i=1;i<=n;++i){
21         while(l<r&&f[q[l+1]]-f[q[l]]<=(S+t[i])*(c[q[l+1]]-c[q[l]]))++l;
22         f[i]=f[q[l]]+S*(c[n]-c[q[l]])+t[i]*(c[i]-c[q[l]]);
23         while(l<r&&(f[i]-f[q[r]])*(c[q[r]]-c[q[r-1]])<=(f[q[r]]-f[q[r-1]])*(c[i]-c[q[r]]))--r;
24         q[++r]=i;
25     }
26     printf("%lld\n",f[n]);
27     return 0;
28 }

3.P3628 特别行动队

没什要说的，斜率的上凸的优化。斜率负的小心点就行了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,1:0;}
 5 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,1:0;}
 6 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 7 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 8 template<typename T>inline T read(T&x){
 9     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c==‘-‘)f=1;
10     while(isdigit(c))x=x*10+c-‘0‘,c=getchar();return f?x=-x:x;
11 }
12 const int N=1000000+7;
13 ll f[N],S[N],a,b,c;
14 int n,l,r,q[N];
15 inline ll fy(int i){return f[i]+a*S[i]*S[i]-b*S[i];}
16 inline ll fx(int i){return S[i];}
17
18 int main(){//freopen("tmp.in","r",stdin);freopen("tmp.out","w",stdout);
19     read(n);read(a),read(b),read(c);q[l=r=1]=0;
20     for(register int i=1;i<=n;++i){
21         S[i]=S[i-1]+read(S[i]);
22         while(l<r&&fy(q[l+1])-fy(q[l])>=2*a*S[i]*(fx(q[l+1])-fx(q[l])))++l;
23         f[i]=f[q[l]]+a*(S[i]-S[q[l]])*(S[i]-S[q[l]])+b*(S[i]-S[q[l]])+c;
24         while(l<r&&(fy(i)-fy(q[r]))*(fx(q[r])-fx(q[r-1]))>=(fy(q[r])-fy(q[r-1]))*(fx(i)-fx(q[r])))--r;
25         q[++r]=i;
26     }
27     printf("%lld\n",f[n]);
28     return 0;
29 }

4.BZOJ3156 防御准备亦无甚要说。longlong注意。状态设计多想想就行,$f_i$表示的是i建塔，从1到i总最小费用，然后转移就简单了。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 typedef double db;
 5 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,1:0;}
 6 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,1:0;}
 7 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 8 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 9 template<typename T>inline T read(T&x){
10     x=0;int f=0;char c;while(!isdigit(c=getchar()))if(c==‘-‘)f=1;
11     while(isdigit(c))x=x*10+(c&15),c=getchar();return f?x=-x:x;
12 }
13 const int N=1000000+7;
14 ll f[N],a[N];
15 int q[N],n,l,r;
16 inline ll y(int j){return f[j]+(1ll*j*(j+1)>>1);}
17
18 int main(){//freopen("tmp.in","r",stdin);freopen("tmp.out","w",stdout);
19     read(n);for(register int i=1;i<=n;++i)read(a[i]);
20     q[l=r=1]=0;
21     for(register int i=1;i<=n;++i){
22         while(l<r&&y(q[l+1])-y(q[l])<=1ll*i*(q[l+1]-q[l]))++l;
23         f[i]=f[q[l]]+(1ll*(i-q[l])*(i-q[l]-1)>>1)+a[i];
24         while(l<r&&(y(i)-y(q[r]))*(q[r]-q[r-1])<=(y(q[r])-y(q[r-1]))*((ll)i-q[r]))--r;
25         q[++r]=i;
26     }
27     printf("%lld\n",f[n]);
28     return 0;
29 }

5.咕咕咕。。

原文地址：https://www.cnblogs.com/saigyouji-yuyuko/p/10389864.html