0/1 score planning topic

0/1 Fractional programming

• For a class ： Each element has two attributes A,B.
• Select several elements , bring $\frac{\sum A_i}{\sum B_i}$ Get the maximum value .
• Usually, the dichotomy test is used , Model as appropriate .
• Monotonicity is required .

Word ring

selected from ：Acwing lew2018 Analysis of

Code ：

#include<bits/stdc++.h>
using namespace std;
#define int long long 
int n;
const int N = 1010,M=100100;
int h[N],e[M],ne[M],idx,w[M];
bool st[N];
int cnt[N];
double dist[N];
void add(int a,int b,int c){
    
    e[idx]=b;
    w[idx]=c;
    ne[idx]=h[a];
    h[a]=idx++;
}
bool check(double x){
    // Average 
    queue<int>q;
    memset(st,0,sizeof st);
    memset(cnt,0,sizeof(cnt));
    for(int i=0;i<676;i++){
    
        q.push(i);
        st[i]=true;
    }
    int count=0;
    while(q.size()){
    
        auto t=q.front();
        q.pop();
        st[t]=false;
        for(int i=h[t];i!=-1;i=ne[i]){
    
            int j=e[i];
            if(dist[j]<dist[t]+w[i]-x){
    // Positive ring 
                dist[j]=dist[t]+w[i]-x;
                cnt[j]=cnt[t]+1;
                // Metaphysical judgment ring 
                if(++count>10000)return true;
                if(cnt[j]>=676)return true;
                if(!st[j]){
    
                    st[j]=true;
                    q.push(j);
                }
            }
        }
    }
    return false;
}
signed main(){
    
    while(cin>>n){
    
        if(!n)break;
        memset(h,-1,sizeof h);
        string s;
        for(int i=1;i<=n;i++){
    
            cin>>s;
            if(s.size()>=2){
    
                int left=(s[0]-'a')*26+s[1]-'a';
                int right=(s[s.size()-2]-'a')*26+s[s.size()-1]-'a';
                int len=s.size();
                add(left,right,len);
            }
        }
        if(!check(0))cout<<"No solution"<<endl;
        else{
    
            double l=0,r=1000;
            while(r-l>=1e-4){
    
                double mid=(r+l)/2;
                if(check(mid))l=mid;
                else r=mid;
            }
            cout<<l<<endl;
        }
    }
    return 0;
}