Submission #412692

#	Time	Username	Problem	Language	Result	Execution time	Memory
412692		losmi247	Boxes with souvenirs (IOI15_boxes)	C++14	100 / 100	619 ms	333012 KiB

boxes

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e7+2;

int n,k,l;
int pos[N];

ll kje1(){
    ll sol = 0;
    for(int i = 0; i < n; i++){
        ll dod = 2*min(pos[i],(int)l-pos[i]);
        sol += dod;
    }
    return sol;
}

ll kjen(){
    int sol = l;
    sol = min(sol,2*(l-pos[0]));
    for(int i = 0; i < n-1; i++){
        int sta = 2*pos[i]+2*(l-pos[i+1]);
        sol = min(sol,sta);
    }
    sol = min(sol,2*pos[n-1]);
    return sol;
}

ll inace(){
    ll sol1 = 0;
    int a = l/2, b = (l+1)/2;
    for(int i = 0; i < n; i += k){
        int kraj = min(i+k-1,n-1);
        /// sad gledas range pos-ova od i do kraj
        if(pos[kraj] <= a) sol1 += (ll)2*pos[kraj];
        else if(pos[i] >= b) sol1 += (ll)2*(l-pos[i]);
        else sol1 += l;
    }

    ll sol2 = 0;
    a = l/2, b = (l+1)/2;
    for(int i = n-1; i >= 0; i -= k){
        int poc = max(0,i-k+1);
        /// sad gledas range pos-ova od poc do i
        if(pos[i] <= a) sol2 += (ll)2*pos[i];
        else if(pos[poc] >= b) sol2 += (ll)2*(l-pos[poc]);
        else sol2 += l;
    }

    return min(sol1,sol2);
}


ll dp1[N],dp2[N];
ll nesto(){
    int a = l/2,b = (l+1)/2;
    for(int i = 0; i < n; i++){
        //if(pos[i] > a) break;
        dp1[i] = 1e18;
        int sta = min(k,i+1);
        dp1[i] = min(dp1[i],2*pos[i]+dp1[i-sta]);
        /*for(int j = 1; j <= k && j <= i+1; j++){
            dp1[i] = min(dp1[i],2*pos[i]+dp1[i-j]);
        }*/
    }
    for(int i = n-1; i >= 0; i--){
        //if(pos[i] < b) break;
        dp2[i] = 1e18;
        int sta = min(k,n-i);
        dp2[i] = min(dp2[i],2*(l-pos[i])+dp2[i+sta]);
        /*for(int j = 1; j <= k && j <= n-i; j++){
            dp2[i] = min(dp2[i],2*(l-pos[i])+dp2[i+j]);
        }*/
    }

    ll sol1 = dp2[0];
    for(int i = 0; i < n; i++) sol1 = min(sol1,dp1[i]+dp2[i+1]);

    ll sol2 = 1e18;
    for(int i = 0; i < n; i++){
        int kraj = min(n-1,i+k-1);
        if(pos[i] <= a && pos[kraj] >= b){
            ll sta = l;
            if(i > 0) sta += dp1[i-1];
            if(kraj < n-1) sta += dp2[kraj+1];
            sol2 = min(sta,sol2);
        }
    }

    return min(sol1,sol2);
}

ll delivery(int br1,int br2,int br3,int *positions){
    n = br1;
    k = br2;
    l = br3;
    for(int i = 0; i < n; i++) pos[i] = positions[i];

    int sta = 0;
    while(sta < n && pos[sta] == 0) sta++;
    if(sta == n) return 0;
    for(int i = 0; i < n-sta; i++) pos[i] = pos[i+sta];
    n -= sta;

    if(k == 1) return kje1();

    if(k >= n) return kjen();

    return nesto();
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 15 / 15
• Subtask 4: 15 / 15
• Subtask 5: 20 / 20
• Subtask 6: 30 / 30