Luogu P1220 turn off street lights

topic

answer:

Easy to prove: At a certain moment, the street lights that are turned off must be continuous.
f[i][j][t] indicates that the street lights of i....j have been turned off, and the current minimum power consumption at i or j (t=0 means at i, t=1 means at j)
Initialization: except f[ Except k][k][0,1]=0, all other f[i][i][t]=inf
state transition equation:
t=s[n]-s[j]+s[i];
f [i][j][0]=min(f[i+1][j][0]+(a[i+1]-a[i])*t,f[i+1][j] [1]+(a[j]-a[i])*t)
t=s[n]-s[j-1]+s[i-1];
f[i][j][1]= min(f[i][j-1][0]+(a[j]-a[i])*t,f[i][j-1][1]+(a[j]-a[ j-1])*t)
t represents the distance

Standard range:

#include<bits/stdc++.h>
using namespace std;
int n,k,i,j,f[53][53][2],s[53],t,a[53],b[53],l;
int main(){
    cin>>n>>k;
    for (i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]);
    for (i=1;i<=n;i++) s[i]=s[i-1]+b[i];
    for (i=1;i<=n;i++) f[i][i][0]=f[i][i][1]=999999999;
    f[k][k][0]=f[k][k][1]=0;
    for (l=2;l<=n;l++)
        for (i=1;i<=n-l+1;i++){
            j=i+l-1;
            t=s[n]-s[j]+s[i];
            f[i][j][0]=min(f[i+1][j][0]+(a[i+1]-a[i])*t,f[i+1][j][1]+(a[j]-a[i])*t);
            t=s[n]-s[j-1]+s[i-1];
            f[i][j][1]=min(f[i][j-1][0]+(a[j]-a[i])*t,f[i][j-1][1]+(a[j]-a[j-1])*t);
        }
    cout<<min(f[1][n][0],f[1][n][1]);
}