[Luogu1220] off street [dynamic programming interval dp]

P1220 off street

An interval dp

Turn off power to the first pre-out i ~ j lamp when

And f [i] [j] [0/1] to light off represents i ~ j minimum cost at this time when the left (0) or right (1)

A bit like [HNOI2010] chorus  this problem is three-dimensional

#include<bits/stdc++.h>
using namespace std;
#define Max(x,y) (x)>(y)?(x):(y)
#define Min(x,y) (x)>(y)?(y):(x)
#define ll long long
#define rg register
const int N=100+5,M=1000000+5,inf=0x3f3f3f3f,P=19650827;
int n,s,sum=0,pos[N],w[N],f[N][N][2],t[N][N];
template <class t>void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}

int main(){
    //freopen("in.txt","r",stdin);
    rd(n),rd(s);
    for(int i=1;i<=n;++i) rd(pos[i]),rd(w[i]),sum+=w[i];
    for(int i=1;i<=n;++i)
    for(int j=i;j<=n;++j) t[i][j]=t[i][j-1]+w[j];
    for(int i=1;i<=n;++i)
    for(int j=i;j<=n;++j) t[i][j]=sum-t[i][j];
    memset(f,inf,sizeof(f));
    f[s][s][0]=f[s][s][1]=0;
    for(int l=2;l<=n;++l)//枚举长度 
    for(int i=1,j;i<=n-l+1;++i){//枚举左端点
        j=l+i-1;
        f[i][j][0]=Min(f[i+1][j][0]+t[i+1][j]*(pos[i+1]-pos[i]),f[i+1][j][1]+t[i+1][j]*(pos[j]-pos[i]));
        f[i][j][1]=Min(f[i][j-1][1]+t[i][j-1]*(pos[j]-pos[j-1]),f[i][j-1][0]+t[i][j-1]*(pos[j]-pos[i]));
    }
    printf("%d",Min(f[1][n][0],f[1][n][1]));
    return 0;
}