【单调队列优化】[CF372C] Watching Fireworks is Fun

突然发现我可能单调队列都打不来了...我太菜了...

这道题显然有$$f[i][j]=min\{f[i-1][k]+\vert j-a[i] \vert\}$$

则$ans=\sum_{i=1}^{m} b_i - min_{j=1}^{n}\{f[m][j]\}$

令$len=(t[i]-t[i-1])*d$则其中k满足$$k∈[j-len,j+len]$$

$f[i][]$只与$f[i-1][]$有关，所以可以把第一维压掉

不难弄出一个$O(n^2m)$的代码：

#include<bits/stdc++.h>
#define int long long
#define writeln(x)  write(x),puts("")
#define writep(x)   write(x),putchar(' ')
using namespace std;
inline int read(){
    int ans=,f=;char chr=getchar();
    while(!isdigit(chr)){if(chr=='-') f=-;chr=getchar();}
    while(isdigit(chr)){ans=(ans<<)+(ans<<)+chr-;chr=getchar();}
    return ans*f;
}void write(int x){
    if(x<) putchar('-'),x=-x;
    if(x>) write(x/);
    putchar(x%+'');
}const int M = ;
int f[][M],n,m,d,ans=0x3f3f3f3f,sum;
struct P{int a,b,t;}a[M];
signed main(){
    n=read(),m=read(),d=read(),sum=;
    for(int i=;i<=m;i++){
        int x=read(),y=read(),z=read();
        sum+=y;
        a[i]=(P){x,y,z};
    }
    for(int i();i<=n;i++)f[][i]=abs(a[].a-i);
    for(int i();i<=m;i++){
        memset(f[i&],0x3f,sizeof(f[i&]));
        int len=(a[i].t-a[i-].t)*d;
        for(int j=;j<=n;j++){
            for(int k=max(1ll,j-len);k<=min(j+len,n);k++)
                f[i&][j]=min(f[i&][j],f[i&^][k]+abs(a[i].a-j));
            if(i==m)ans=min(ans,f[i&][j]);
        }
    }
    cout<<sum-ans<<endl;
    return ;
}

但是显然会T，注意到上面k的范围，可以考虑单调队列优化，对于每一次的$i,j$，对$[j-len,j]$跑一次滑动窗口，对$[j,j+len]$跑一次滑动窗口取最小值即可

 #include<bits/stdc++.h>
 #define int long long
 #define writeln(x)  write(x),puts("")
 #define writep(x)   write(x),putchar(' ')
 using namespace std;
 inline int read(){
     int ans=,f=;char chr=getchar();
     while(!isdigit(chr)){if(chr=='-') f=-;chr=getchar();}
     while(isdigit(chr)){ans=(ans<<)+(ans<<)+chr-;chr=getchar();}
     return ans*f;
 }void write(int x){
     if(x<) putchar('-'),x=-x;
     if(x>) write(x/);
     putchar(x%+'');
 }const int M = ;
 int f[][M],n,m,d,ans=0x3f3f3f3f,sum,q[M];
 struct P{int a,b,t;}a[M];
 signed main(){
     n=read(),m=read(),d=read(),sum=;
     for(int i=;i<=m;i++){
         int x=read(),y=read(),z=read();
         sum+=y;a[i]=(P){x,y,z};
     }for(int i();i<=n;i++)f[][i]=abs(a[].a-i);
     for(int i();i<=m;i++){
         int len=(a[i].t-a[i-].t)*d,l=,r=;
         for(int j=;j<=n;j++){
             while(l<=r&&q[l]<j-len)++l;
             while(l<=r&&f[i&^][q[r]]>f[i&^][j])--r;
             q[++r]=j;
             f[i&][j]=f[i&^][q[l]]+abs(j-a[i].a);
             if(i==m)ans=min(ans,f[i&][j]);
         }l=,r=;
         for(int j=n;j>=;j--){
             while(l<=r&&q[l]-len>j)++l;
             while(l<=r&&f[i&^][q[r]]>f[i&^][j])--r;
             q[++r]=j;
             f[i&][j]=min(f[i&][j],f[i&^][q[l]]+abs(j-a[i].a));
             if(i==m)ans=min(ans,f[i&][j]);
         }
     }cout<<sum-ans<<endl;
     return ;
 }