Submission #636634

#	Time	Username	Problem	Language	Result	Execution time	Memory
636634		gromperen	Aliens (IOI16_aliens)	C++14	0 / 100	1 ms	340 KiB

aliens

    #include <bits/stdc++.h>
    #include "aliens.h"


    #define ll long long
    #define int long long
    using namespace std;


    const int INF = 1e18;


    int sqr(int x) {return x * x;}

    long long take_photos(int32_t n, int32_t m, int32_t k, vector<int32_t> r, vector<int32_t> c) {

    	vector<pair<int,int>> tmpcords;
    	vector<pair<int,int>> cords;
    	for (int i = 0; i < n; ++i) {
    		tmpcords.push_back({min(r[i], c[i]), max(r[i], c[i])});
    	}
    	sort(tmpcords.begin(), tmpcords.end(),
    			[](pair<int,int>a,pair<int,int>b)
    			{return a.first<b.first || (a.first==b.first && a.second>b.second);});

    	for (auto i : tmpcords) {
    		if (cords.size() && cords.back().second >= i.second) continue;
    		cords.push_back(i);
    	}

    	n = cords.size();
	k = min(k, n);
    	//cout << "DBG" << endl;
    	//for (int i =0 ;i < n; ++i) {
    		//cout << cords[i].first << " " << cords[i].second << "\n";

    	//}



	auto cost = [&](int i, int j) {
		// area of photo - overlap
		if (i == 0)
			return sqr(cords[j].second - cords[i].first + 1);
		else
			return sqr(cords[j].second - cords[i].first + 1) - max(0LL, sqr(cords[i-1].second - cords[i].first + 1));

	};

    	vector<vector<int>> dp(n+5, vector<int>(k+5, INF));
    	vector<vector<int>> opt(n+5, vector<int>(k+5, INF));
	for (int i = 0; i < n; ++i) {
		//opt[i][i] = i; // if you have n == k then optimum is to take picture of every segment
		opt[i][1] = 0; // if you have 1 picture optimum is to start at 0
		dp[i][1] = cost(0, i);
	}
    	//dp[0][0] = 0;
    	//for (int i = 0;i <= k; ++i) dp[0][i] = 0;

    	for (int i = n-1; i >= 0; --i) {
    		for (int j =2; j <= k; ++j) {
			opt[n][j] = n-2;
			dp[i][j] = dp[i][j-1];
    			for (int p = opt[i][j-1]; p <= opt[i+1][j]; ++p) {
				int cur = dp[p][j-1] + cost(p+1, i);
    				//int cur = dp[p][j-1] + sqr(cords[i-1].second - cords[p].first + 1);
    				//if ((p -1 >= 0) && cords[p-1].second >= cords[p].first)
    					//cur -= sqr(cords[p-1].second - cords[p].first + 1);
				if (cur < dp[i][j]) {
					opt[i][j] = p;
					dp[i][j] = cur;
				}
    				//cout << cur << " ";
    				//cout << cur << endl;
    				 //= min(dp[i][j], cur);
    			}
    		}
    	}

    	//for (int i =0; i<=n;++i) {
    		//for (int j = 0; j<=k;++j) {
    			//cout << dp[i][j] << " ";
    		//}
    		//cout << "\n";
    	//}

    	int ans = INF;
    	for (int i = 0; i <= k; ++i) {
    		ans = min(ans, dp[n-1][i]);
    	}


        return ans;
    }

• Subtask 1: 0 / 4
• Subtask 2: 0 / 12
• Subtask 3: 0 / 9
• Subtask 4: 0 / 16
• Subtask 5: 0 / 19
• Subtask 6: 0 / 40