oj.uz

Submission #302405

# Submission time Handle Problem Language Result Execution time Memory
302405 2020-09-18T16:23:27 Z user202729 Palembang Bridges (APIO15_bridge) C++17
78 / 100
 1082 ms 17124 KB 
// might be fast enough. Still log^terrible.



#if not LOCAL
//#define NDEBUG 1
#endif
#include<bits/stdc++.h>

std::vector<int> positions; // it's more convenient to make it global. It's not changed very frequently anyway

/*
 * stores a convex piecewise linear function.
 * the cuts must be in positions.
 */

struct Tree{
	using Data=std::array<int64_t, 2>; // sort of parallel data tuple. Can be used for anything.
	std::vector<Data> data; // "lazy"
	Tree(std::size_t number): data(number*2){}

	static void add1(Data& first, Data sec){
		first[0]+=sec[0];
		first[1]+=sec[1];
	}

	int offset() const{return int(data.size()/2);}

	// get raw {a, b} value at a particular position.
	Data rawGet(int index)const{
		index+=offset();
		Data result{};
		for(; index; index>>=1)
			add1(result, data[index]);
		return result;
	}

	// add a value to a range.
	void add(int left, int right, Data const value){
		left+=offset(); right+=offset();
		//assert(left<=right); // okay
		while(left<right){
			if(left&1) add1(data[left++], value);
			if(right&1) add1(data[--right], value);
			left>>=1; right>>=1;
		}
	}

	// self explanatory.
	int64_t getAtIndex(int index)const{
		auto const [a, b]=(*this).rawGet(index);
		return (int64_t)a*positions[index]+b;
	}

	// get the function value, at a particular (possibly not in positions) position.
	int64_t getAtValue(int pos)const{


		// interpolate if necessary. (always work in this particular problem...)
		// could be a little simpler (by explicitly store the slope on each segment), but...
		// :(

		auto const iterator=std::lower_bound(begin(positions), end(positions), pos);
		//assert(iterator!=positions.end()); // all passed
		int64_t const value2=getAtIndex(int(iterator-positions.begin()));
		if(*iterator==pos) return value2;
		//assert(iterator!=positions.begin());
		//assert(iterator[-1]<pos); assert(pos<iterator[0]);
		int64_t const value1=getAtIndex(int(iterator-positions.begin())-1);

		auto const number=value2-value1;
		auto const de=*iterator-iterator[-1];
		assert(number%de==0);
		return number/de*(pos-iterator[-1])+value1;
	}

	// get any index for which getAtIndex returns the minimum value.
	int minIndex() const{

		// assumes some specific kind of function shape.
		// currently O(log^2 n)
		// could be O(log n) instead.
		// :(

		// the boundary binary search might be reducible to O(log n)
		// too with some careful planning. I don't know.

		auto const value=[&](int pos){
			//assert(0<=pos and pos<offset()); // okay
			return (*this).rawGet(pos)[0];
		};
		int pos=offset(); // relatively small (compared to INT_MAX)
		for(int step=1<<30; step>>=1;){
			if(pos-step>=0 and value(pos-step)>=0) pos-=step;
		}
		// now pos=first element with value >=0

		std::pair<int64_t, int> result{INT64_MAX, INT_MAX};
		for(int pos1=pos-1; pos1<=pos; ++pos1){ // perhaps this can be simpler? I don't know...
			if((unsigned) pos1<(unsigned)offset())
				result=std::min(result, std::make_pair(getAtIndex(pos1), pos1));
		}

		/*
#warning
		[&]{ // (slow) ensure that the found index is really the minimum.
			for(int pos1=0; pos1<offset(); ++pos1){
				if(not(getAtIndex(pos1)>=result.first)) std::exit(1);
			}
			return true;
		}();
		*/
		return result.second;
	}
};

int main(){
	std::ios::sync_with_stdio(0);std::cin.tie(0);
	int k, number; std::cin>>k>>number;
	if(k==1) return 1;
	int64_t result{};
	std::vector<std::array<int, 2>> queries;
	// read into queries, and prepare result (nonoptimizable part).
	// this part should not be wrong.
	for(int _=0; _<number; ++_){
		char a, b; int first, sec;
		std::cin>>a>>first>>b>>sec;
		if(first>sec) std::swap(first, sec);
		result+=sec-first; // minimum required
		if(a!=b){
			result+=1; // need to cross a bridge anyway
			queries.push_back({first, sec});
			positions.push_back(first); positions.push_back(sec);
		}
	}
	number=-1;

	// sort positions.
	std::sort(begin(positions), end(positions));
	positions.erase(std::unique(begin(positions), end(positions)), end(positions));
	// from now on, positions are not changed anymore.

	if(positions.size()<=2){
		std::cout<<result<<'\n';
		return 0;
	}

	auto minAdd=INT64_MAX;
	// minimum of the function (it's hard to explain.)

	std::sort(begin(queries), end(queries), [&](auto const& first, auto const& sec){
		return first[0]+first[1]<sec[0]+sec[1];
	});

	Tree ffirst(positions.size()), fsec(positions.size());
	// in this problem value (a, b) means y=a*x+b

	/*
	first and sec are values in positions.
	Add the function multiplier*max(first-x, 0, x-sec) to tree.
	(it's convex if multiplier>=0.)
	*/
	auto const addSlope=[&](Tree& tree, int first, int sec, int multiplier){
		//assert(first<=sec); // okay
		auto const ifirst=int(std::lower_bound(begin(positions), end(positions), first)-begin(positions));
		auto const isec=int(std::lower_bound(begin(positions), end(positions), sec)-begin(positions));
		//assert(positions[ifirst]==first); // okay
		//assert(positions[isec]==sec);
		//assert(tree.offset()==(int)positions.size()); // okay
		tree.add(0, ifirst, {-1*multiplier, first*multiplier});
		tree.add(isec, tree.offset(), {1*multiplier, -sec*multiplier});
		// * Data could be std::pair<int, int64_t> * save implementation time!
	};
	for(auto [first, sec]: queries)
		addSlope(fsec, first, sec, 1);

	for(int slice=0, lastSum=0; slice<=(int)queries.size(); ++slice){
		int curSum=slice==(int)queries.size() ? positions.back()*2: queries[slice][0]+queries[slice][1];
		// 2e9'ish
		if(curSum>lastSum){
			// find minAdd in the current slice (first<=sec, lastSum<=first+sec<=curSum)

			auto const searchFunction=[&](auto function, int left, int right){
				// minAdd=min(minAdd, function(i)) for i in left..=right
				// given that function is ternary (or something like that) on the segment
				//
				// complexity: log(value) calls to function. (which is actually pretty terrible?)

				if(left>right) return; // just to be sure

				int pos=left; auto value=function(pos);
				auto const process=[&](int pos1){
					auto const value1=function(pos1);
					if(value1<value){
						pos=pos1; value=value1;
					}
				};
				for(int step=1<<30; step; step>>=1){
					if((int64_t)pos-step>=left) process(pos-step);
					if((int64_t)pos+step<=right) process(pos+step);
				}
				minAdd=std::min(minAdd, value);

				//assert(value==function(pos)); // okay
				//assert(left<=pos); assert(pos<=right);
				/*
				assert(right-left<=500); // * only while debugging
				for(int i=left; i<=right; ++i)
					assert(function(i)>=value);
					*/
			};
			auto const searchSlope=[&](int sum){
				//assert(lastSum<=sum); assert(sum<=curSum); // okay
				//assert(sum>=0); // okay, implied by previous

				// diff -> (sum-diff)/2, (sum+diff)/2
				// diff=sum&1+2*x
				// => ((sum-sum&1-2x)/2, (sum+sum&1+2x)/2)
				// => (  (sum>>1)-x,  ((sum+1)>>1)+x  )
				// obviously first<=sec
				// need positions[0]<=first, sec<=positions.back()

				// (to clarify: it's not "necessary", the result remains correct without that,
				// but the current implementation of Tree::getAtValue can only handle those values)
				searchFunction([&](int x){
					auto const first=(sum>>1)-x, sec=((sum+1)>>1)+x;
					//assert(x>=0); // looks good
					//assert(first<=sec); // okay, basic arithmetic and implied by above
					//assert(first+sec==sum); // ^

					return ffirst.getAtValue(first)+fsec.getAtValue(sec);
				}, 0, std::min(positions.back()-((sum+1)>>1), (sum>>1)-positions[0]));
			};

			auto const searchC=[&]{
				searchFunction([&](int value){
					//assert(lastSum<=value+value); // okay
					//assert(value+value<=curSum);

					return ffirst.getAtValue(value)+fsec.getAtValue(value);
				}, std::max(positions[0], (lastSum+1)>>1), std::min(positions.back(), curSum>>1));
			};

			auto const ifirst=ffirst.minIndex();
			auto const isec=fsec.minIndex();
			auto const sum1=positions[ifirst]+positions[isec]; // range: 2e9'ish

			auto const minValue=ffirst.getAtIndex(ifirst)+fsec.getAtIndex(isec);
			if(minValue<minAdd){
				if(lastSum<=sum1 and sum1<=curSum and ifirst<=isec){
					minAdd=std::min(minAdd, minValue);
				}else{ // minValue is not reachable. Only larger values are.
					searchC();
					if(sum1<lastSum)
						searchSlope(lastSum);
					if(curSum<sum1)
						searchSlope(curSum);
				}
			}

			lastSum=curSum;
		}else{
			//assert(curSum==lastSum); // okay
		}

		if(slice!=(int)queries.size()){
			// apply change from queries[slice]
			auto const [first, sec]=queries[slice];
			addSlope(fsec, first, sec, -1);
			addSlope(ffirst, first, sec, 1);
		}
	}

	std::cout<<result+minAdd*2<<'\n';
}

Subtask #1
0 / 8.0

# Verdict Execution time Memory Grader output
1 Runtime error 1 ms 384 KB Execution failed because the return code was nonzero
2 Halted 0 ms 0 KB -

Subtask #2
0 / 14.0

# Verdict Execution time Memory Grader output
1 Runtime error 0 ms 384 KB Execution failed because the return code was nonzero
2 Halted 0 ms 0 KB -

Subtask #3
9.0 / 9.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 288 KB Output is correct
2 Correct 1 ms 512 KB Output is correct
3 Correct 1 ms 384 KB Output is correct
4 Correct 1 ms 384 KB Output is correct
5 Correct 1 ms 384 KB Output is correct
6 Correct 1 ms 384 KB Output is correct
7 Correct 1 ms 384 KB Output is correct
8 Correct 1 ms 384 KB Output is correct
9 Correct 1 ms 384 KB Output is correct
10 Correct 1 ms 384 KB Output is correct
11 Correct 1 ms 384 KB Output is correct
12 Correct 1 ms 384 KB Output is correct

Subtask #4
32.0 / 32.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 384 KB Output is correct
2 Correct 1 ms 384 KB Output is correct
3 Correct 1 ms 384 KB Output is correct
4 Correct 1 ms 384 KB Output is correct
5 Correct 1 ms 384 KB Output is correct
6 Correct 1 ms 384 KB Output is correct
7 Correct 1 ms 384 KB Output is correct
8 Correct 1 ms 384 KB Output is correct
9 Correct 1 ms 384 KB Output is correct
10 Correct 1 ms 384 KB Output is correct
11 Correct 1 ms 384 KB Output is correct
12 Correct 1 ms 512 KB Output is correct
13 Correct 1 ms 384 KB Output is correct
14 Correct 2 ms 384 KB Output is correct
15 Correct 7 ms 512 KB Output is correct
16 Correct 1 ms 384 KB Output is correct
17 Correct 1 ms 384 KB Output is correct
18 Correct 3 ms 384 KB Output is correct
19 Correct 1 ms 416 KB Output is correct
20 Correct 7 ms 512 KB Output is correct
21 Correct 2 ms 512 KB Output is correct
22 Correct 7 ms 512 KB Output is correct
23 Correct 1 ms 384 KB Output is correct
24 Correct 4 ms 512 KB Output is correct

Subtask #5
37.0 / 37.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 384 KB Output is correct
2 Correct 0 ms 384 KB Output is correct
3 Correct 1 ms 512 KB Output is correct
4 Correct 1 ms 512 KB Output is correct
5 Correct 1 ms 384 KB Output is correct
6 Correct 1 ms 384 KB Output is correct
7 Correct 1 ms 384 KB Output is correct
8 Correct 1 ms 384 KB Output is correct
9 Correct 1 ms 512 KB Output is correct
10 Correct 1 ms 384 KB Output is correct
11 Correct 1 ms 384 KB Output is correct
12 Correct 1 ms 384 KB Output is correct
13 Correct 1 ms 384 KB Output is correct
14 Correct 2 ms 416 KB Output is correct
15 Correct 7 ms 640 KB Output is correct
16 Correct 1 ms 384 KB Output is correct
17 Correct 1 ms 384 KB Output is correct
18 Correct 2 ms 384 KB Output is correct
19 Correct 1 ms 384 KB Output is correct
20 Correct 7 ms 512 KB Output is correct
21 Correct 2 ms 512 KB Output is correct
22 Correct 7 ms 640 KB Output is correct
23 Correct 1 ms 384 KB Output is correct
24 Correct 3 ms 512 KB Output is correct
25 Correct 24 ms 2288 KB Output is correct
26 Correct 61 ms 2292 KB Output is correct
27 Correct 784 ms 13668 KB Output is correct
28 Correct 1082 ms 14944 KB Output is correct
29 Correct 1074 ms 17124 KB Output is correct
30 Correct 531 ms 9316 KB Output is correct
31 Correct 33 ms 3816 KB Output is correct
32 Correct 953 ms 17124 KB Output is correct
33 Correct 127 ms 16608 KB Output is correct
34 Correct 951 ms 16468 KB Output is correct
35 Correct 41 ms 4064 KB Output is correct
36 Correct 403 ms 16868 KB Output is correct
37 Correct 45 ms 2932 KB Output is correct