Fence
\(K\)很小，考虑\(K\)开一维，\(N\)开一维
\(f_{i,j}\)表示前\(i\)个工匠粉刷前\(j\)个木板的最大花费
\(f_{i,j}=\min_{k=j-l_i}^{s_i-1} f_{i-1,k}+(j-k) \cdot p_i\)
拆开为
\(f_{i,j}=f_{i-1,k}-k \cdot p_i+j \cdot p_i\)
\(i\)固定时维护\(f_{i-1,k}-k \cdot p_i\)的最小值即可
应该维护一个从\(s_i-1\)开始的后缀min，单调队列也行，但数组更香
PS:数据范围是不是错了，调了一年
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define fr(i,a,b) for(int i=a;i<=b;i++)
#define N 105
#define M 160005
#define ll long long
int n,m,L[N],P[N],S[N];
ll mn[M],f[N][M];
struct node{int l,p,s;
}a[N];
bool cmp(node A,node B){return A.s<B.s;
}
int main(){scanf("%d%d",&m,&n);fr(i,1,n)scanf("%d%d%d",&a[i].l,&a[i].p,&a[i].s);sort(a+1,a+n+1,cmp);fr(i,1,n)L[i]=a[i].l,P[i]=a[i].p,S[i]=a[i].s;fr(i,1,n){memset(mn,-0x3f,sizeof mn);for(int j=S[i]-1;j>=0;j--)mn[j]=max(mn[j+1],f[i-1][j]-1ll*j*P[i]);fr(j,S[i],S[i]+L[i]-1)f[i][j]=mn[max(0,j-L[i])]+1ll*j*P[i];fr(j,1,m)f[i][j]=max(f[i][j],max(f[i-1][j],f[i][j-1]));}printf("%lld\n",f[n][m]);return 0;
}
P2569 [SCOI2010]股票交易
其实不难，嘴巴AC算了
\(f_{i,j}\)表示第\(i\)天持有\(j\)股，当前收益max值
然后就是大分讨，拆式子，单调队列
#include<bits/stdc++.h>
using namespace std;
#define fr(i,a,b) for(int i=a;i<=b;i++)
#define N 2021
#define inf 0x3f3f3f3f
int n,m,k,ap[N],bp[N],as[N],bs[N],f[N][N];
deque<int> q;
int main(){scanf("%d%d%d",&n,&m,&k);fr(i,1,n)scanf("%d%d%d%d",&ap[i],&bp[i],&as[i],&bs[i]);memset(f,-0x3f,sizeof(f));fr(i,1,n){fr(j,0,as[i])f[i][j]=-j*ap[i];fr(j,0,m)f[i][j]=max(f[i][j],f[i-1][j]);if(i<=k)continue;q.clear();fr(j,0,m){while(q.size()&&j-q.front()>as[i])q.pop_front();while(q.size()&&f[i-k-1][j]+j*ap[i]>=f[i-k-1][q.back()]+q.back()*ap[i])q.pop_back();q.push_back(j);if(q.size())f[i][j]=max(f[i][j],f[i-k-1][q.front()]+(q.front()-j)*ap[i]);}q.clear();for(int j=m;j>=0;j--){while(q.size()&&q.front()-j>bs[i])q.pop_front();while(q.size()&&f[i-k-1][j]+j*bp[i]>=f[i-k-1][q.back()]+q.back()*bp[i])q.pop_back();q.push_back(j);if(q.size())f[i][j]=max(f[i][j],f[i-k-1][q.front()]+(q.front()-j)*bp[i]);}}printf("%d",f[n][0]);return 0;
}
P4954 [USACO09OPEN]Tower of Hay G
题目大意：\(n\)个数划分成若干个连续段，前一段sum\(\geq\)后一段sum，问最多划分几段
一个明显的DP是:
\(f_{i,j}\)表示前\(i\)个,最后一段为\(i\)~\(j\)的最高层数
但很遗憾这样最少也是\(n^2\)无法再优化了
考虑干掉第二维
手玩几组样例后发现，最优解一定是可行解中底部最小的
直觉上就是传上去了，不亏
证明用数归即可
因而只需确认底层最少几个，再递归即可
但还是难转移，因为底层看不到上面
所以只能从上面看下来
因此倒做
考虑\(f_i\)表示\(i\)~\(n\)分好的最高层数
\(g_i\)表示此时底层最小宽度
则\(f_i=\max_{j=i+1}^n f_j+1\) if \(s_{i...j-1} \geq g_j\)
\(O(n^2)\)
但是if后的条件\(s_{j-1}-s_{i-1} \geq g_j \leftrightarrow s_{i-1} \leq s_{j-1}-g_j\)
为了好看可改为后缀和
\(s_i-s_j \geq g_j\) , \(s_i \geq g_j+s_j\)
不难看出单调性，上优化即可
不得不说是个好题，值得一讲
#include<bits/stdc++.h>
using namespace std;
#define fr(i,a,b) for(int i=a;i<=b;i++)
#define N 100005
int n,a[N],s[N],f[N],g[N];
int q[N],hh,tt;
int main(){scanf("%d",&n);fr(i,1,n)scanf("%d",&a[n-i+1]);fr(i,1,n)s[i]=s[i-1]+a[i];fr(i,1,n){while(hh<=tt&&s[q[hh]]+g[q[hh]]<=s[i])hh++;g[i]=s[i]-s[q[hh-1]],f[i]=f[q[hh-1]]+1;while(hh<=tt&&s[q[tt]]+g[q[tt]]>=s[i]+g[i])tt--;q[++tt]=i;}printf("%d\n",f[n]);return 0;
}
P5665 [CSP-S2019] 划分
一模一样的题。。。
突然发现家里电脑没有int128？？？
空间实在恶心，88pts算了
#include<bits/stdc++.h>
using namespace std;
#define fr(i,a,b) for(int i=a;i<=b;i++)
#define N 40000005
#define ll long long
#define wow __int128
int n,typ,b1,b2,x,y,z,m,mod=(1<<30);
ll s[N],g[N];
wow f[N];
int q[N],hh,tt;
void write(wow x){if(x>(wow)9)write(x/10);putchar(x%10+'0');
}
int get(int i){if(i<=2)return i==1?b1:b2;else{int ans=(1ll*b2*x+1ll*b1*y+1ll*z)%mod;b1=b2,b2=ans;return ans;}
}
int main(){scanf("%d%d",&n,&typ);if(typ==0){fr(i,1,n){scanf("%d",&b1);s[i]=s[i-1]+b1;}}else{scanf("%d%d%d%d%d%d",&x,&y,&z,&b1,&b2,&m);int L=1,R,l,r;fr(i,1,m){scanf("%d%d%d",&R,&l,&r);fr(j,L,R)s[j]=s[j-1]+(get(j)%(r-l+1))+l;L=R+1;}}fr(i,1,n){while(hh<=tt&&s[q[hh]]+g[q[hh]]<=s[i])hh++;hh--;g[i]=s[i]-s[q[hh]],f[i]=f[q[hh]]+(wow)g[i]*g[i];while(hh<=tt&&s[q[tt]]+g[q[tt]]>=s[i]+g[i])tt--;q[++tt]=i;}write(f[n]);return 0;
}
哦，看到某位大神题解，对于压空间深受启发，AC code：
#include<bits/stdc++.h>
using namespace std;
#define fr(i,a,b) for(int i=a;i<=b;i++)
#define N 40000005
#define ll long long
#define wow __int128
int n,typ,b1,b2,x,y,z,m,mod=(1<<30);
ll s[N],g[N];
struct node{int id;wow f;
};
deque<node> q;
void write(wow x){if(x>(wow)9)write(x/10);putchar(x%10+'0');
}
int get(int i){if(i<=2)return i==1?b1:b2;else{int ans=(1ll*b2*x+1ll*b1*y+1ll*z)%mod;b1=b2,b2=ans;return ans;}
}
int main(){scanf("%d%d",&n,&typ);if(typ==0){fr(i,1,n){scanf("%d",&b1);s[i]=s[i-1]+b1;}}else{scanf("%d%d%d%d%d%d",&x,&y,&z,&b1,&b2,&m);int L=1,R,l,r;fr(i,1,m){scanf("%d%d%d",&R,&l,&r);fr(j,L,R)s[j]=s[j-1]+(get(j)%(r-l+1))+l;L=R+1;}}node hh,tt;q.push_back((node){0,0});fr(i,1,n){while(q.size()&&s[q.front().id]+g[q.front().id]<=s[i])hh=q.front(),q.pop_front();q.push_front(hh);tt.id=i;g[i]=s[i]-s[hh.id],tt.f=hh.f+(wow)g[i]*g[i];while(q.size()&&s[q.back().id]+g[q.back().id]>=s[i]+g[i])q.pop_back();q.push_back(tt);
// puts("orz");}write(q.back().f);return 0;
}
acwing 299.裁剪序列
很有趣的一个题
首先还是无脑DP，设\(f_i\)表示\(1\)到\(i\)分段最小代价
则\(f_i = f_j + maxa(i\)~\(j)\)
st表维护最值，\(O(n^2)\)转移即可
考虑优化，发现一段区间在满足和限制且最大值不变的情况下，一定会尽量扩张
因此可以找出对于\(i\)向左，使最大值改变的所有位置，从左到右记为\(p_1,p_2,...p_k\)
不难看出其单调递减，单调队列即可
\(p_i\)贡献为\(f_{p_i}+p_{i+1}\)
拿个堆维护一下就行
当然还有和限制，特判即可
#include<bits/stdc++.h>
using namespace std;
#define fr(i,a,b) for(int i=a;i<=b;i++)
#define N 100005
#define ll long long
int n,a[N],vis[N];
ll m,f[N];
int q[N],hh,tt;
struct node{int fo,la;ll d;friend bool operator < (node A,node B){return A.d>B.d;}
};
priority_queue<node> p;
int main(){scanf("%d%lld",&n,&m);fr(i,1,n){scanf("%d",&a[i]);if(a[i]>m){puts("-1");return 0;}}ll sum=0;int j=0;fr(i,1,n){sum+=a[i];while(sum>m)sum-=a[++j];while(hh<=tt&&q[hh]<j)vis[q[hh++]]=1;while(hh<=tt&&a[i]>=a[q[tt]])vis[q[tt--]]=1;if(hh<=tt)p.push((node){q[tt],i,f[q[tt]]+a[i]});q[++tt]=i;f[i]=f[j]+a[q[hh]];p.push((node){q[tt],i,f[q[tt]]+a[i]});while(p.size()&&(vis[p.top().fo]||vis[p.top().la]))p.pop();if(p.size())f[i]=min(f[i],p.top().d);}printf("%lld\n",f[n]);return 0;
}
总结：状态尽量设计的好看一些，分离各个变量，就能用单调队列了
