Submission #254883

#	Time	Username	Problem	Language	Result	Execution time	Memory
254883		Hehehe	Wiring (IOI17_wiring)	C++14	100 / 100	109 ms	18280 KiB

wiring

#include<bits/stdc++.h> //:3
using namespace std;
typedef long long ll;
#define all(a) (a).begin(), (a).end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define rc(s) return cout<<s,0
#define pi pair <ll, ll>
#define sz(x) (int)((x).size())


const int N = 2e6 + 11;
const ll INF64 = 3e18 + 1;


ll min_total_length(vector<int> red, vector<int> blue){
    
    vector<ll>a, b;
    vector<pi>v;
    
    for(auto it : red){
        v.pb({it, 0});
    }
    
    for(auto it : blue){
        v.pb({it, 1});
    }
    
    sort(all(v));
    
    for(auto it : v){
        b.push_back(it.ff);
        a.push_back(it.ss);
    }
    
    vector<ll>close(sz(v) + 5, 0);
    vector<ll>match(sz(v) + 5, -1);
    vector<ll>matchval(sz(v) + 5, 0);
    
    int n = sz(v);
    
    for(int i = 0; i < n; i++){
        int j = i;
        
        while (j + 1 < n && a[j + 1] == a[i])j++; //block of type a[i]
        
        for (int k = i; k <= j; k++){
            close[k] = min(i ? b[k] - b[i - 1] : INF64, j + 1 < n ? b[j + 1] - b[k] : INF64);
            //close[k] - distance from the closest block of different color 
        }
            
        i = j;
    }
    
    for(int i = 1; i < n; i++){
        if(a[i] == a[i - 1])continue;
        
        for(int l = i - 1, r = i; (l >= 0 && r < n && a[l] == a[i - 1] && a[r] == a[i]); l--, r++){
                
            match[r] = l;
            matchval[r] = b[r] - b[l];
            
            if(r > i){
                matchval[r] += matchval[r - 1]; //if the block has the size larger than 1
            }
        }
    }
    
    vector<ll>dp(n + 5, 0);
    
    dp[0] = close[0];
    for(int i = 1; i < n; i++){
        
        dp[i] = dp[i - 1] + close[i];
        
        if(match[i] == -1)continue;
        
        ll DP = (match[i] ? dp[match[i] - 1] : 0);
        DP += matchval[i];
        
        dp[i] = min(dp[i], DP);
        
    }
    
    return dp[n - 1];
}

• Subtask 1: 7 / 7
• Subtask 2: 13 / 13
• Subtask 3: 10 / 10
• Subtask 4: 25 / 25
• Subtask 5: 45 / 45