제출 #784736

#제출 시각아이디문제언어결과실행 시간메모리
784736YassirSalamaShortcut (IOI16_shortcut)C++14
0 / 100
1 ms340 KiB
#include "shortcut.h" #include<bits/stdc++.h> using namespace std; #define ll long long const ll MAXN=5100; const ll INF=1e9+10; #define dbg(x) cout<<" [ "<<#x<<" ] : "<<x<<endl; #define F first #define S second #define all(v) v.begin(),v.end() long long find_shortcut(int n, vector<int> l, vector<int> d, int c) { if(n==2){ if(c<l[0]) return d[0]+c+d[1]; return d[0]+l[0]+d[1]; } vector<int> pref(MAXN+1,0); for(int i=0;i<n-1;i++){ pref[i+1]=pref[i]+l[i]; } vector<int> left(MAXN+1,0); vector<int> right(MAXN+1,0); //the idea is to use pref[i-1]+l[i] or d[i] left[0]=d[0]; int first_node=0; for(int i=1;i<n;i++){ if(d[i]>left[i-1]+l[i-1]) first_node=i; left[i]=max(left[i-1]+l[i-1],d[i]); } int last_node=n-1; right[n-1]=d[n-1]; for(int i=n-2;i>=0;i--){ if(d[i]>right[i+1]+l[i]) last_node=i; right[i]=max(right[i+1]+l[i],d[i]); } vector<int> nodes;//nodes in the diameter // cout<<first_node<<" "<<last_node<<endl; int diameter=left[first_node]+right[first_node]; // cout<<diameter<<endl; for(int i=first_node;i<=last_node;i++){ for(int j=i+1;j<=last_node;j++){ int express=left[i]+c+right[j]; int newdiameter=0; for(int k=0;k<=n-1;k++){ newdiameter=max(newdiameter,abs(pref[j]-pref[k])+right[j]); } for(int k=n-1;k>=0;k--){ newdiameter=max(newdiameter,abs(pref[k]-pref[i])+left[i]); } // if(i==0&&j==7){ // cout<<pref[j]- // } // cout<<newdiameter<<" "<<express<<" "<<i<<" "<<j<<endl; // if(newdiameter>express) continue; diameter=min({max(newdiameter,express),diameter}); } } // cout<<c<<endl; return diameter; } // int main() // { // int n, c; // assert(2 == scanf("%d%d", &n, &c)); // std::vector<int> l(n - 1); // std::vector<int> d(n); // for (int i = 0; i < n - 1; i++) // assert(1 == scanf("%d", &l[i])); // for (int i = 0; i < n; i++) // assert(1 == scanf("%d", &d[i])); // long long t = find_shortcut(n, l, d, c); // printf("%lld\n", t); // return 0; // }
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