决策单调性 + WQS二分

贴个代码先...

//by Judge#pragma GCC optimize("Ofast")#include
    
     #define Rg register#define fp(i,a,b) for(Rg int i=(a),I=(b)+1;i
     
      I;--i)#define ll long longusing namespace std;const int M=4003;#ifndef Judge#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)#endifchar buf[1<<21],*p1=buf,*p2=buf;inline ll read(){ ll x=0,f=1; char c=getchar();    for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;    for(;isdigit(c);c=getchar()) x=x*10+c-'0'; return x*f;} int n,K,ans,s[M][M],f[M],w[M];inline int calc(int j,int i){    return f[j]+s[i][i]-s[j][i];}inline bool judge(int j,int k,int i){ //判断 f[i] 大小     int valj=calc(j,i),valk=calc(k,i);    if(valj^valk) return valj>valk;    return w[j]>=w[k];}inline int rate(int j,int k){  //得到交点位置     int l=k+1,r=n;    while(l<=r){        int mid=(l+r)>>1;        if(judge(j,k,mid)) r=mid-1;        else l=mid+1;    } return l;}inline bool check(int mid){ //二分附加权值     static int head,tail,q[M];    q[head=tail=1]=0;    fp(i,1,n){ //斜率优化         while(head
      
       rate(q[tail],i)) --tail; q[++tail]=i;    } return w[n]<=K;}int main(){ n=read(),K=read();    fp(i,1,n) fp(j,1,n) s[i][j]=read();    fp(i,1,n) fp(j,1,i) s[i][j]=0;    fp(i,1,n) fp(j,1,n) s[i][j]=s[i][j-1]+s[i][j];    fp(i,1,n) fp(j,1,n) s[i][j]=s[i-1][j]+s[i][j];    int l=0,r=s[n][n];    while(l<=r){ int mid=(l+r)>>1;        if(check(mid)) r=mid-1,ans=f[n]-K*mid;        else l=mid+1;    } return !printf("%d\n",ans);}