답안 #319090

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
319090 2020-11-03T23:19:08 Z AmineWeslati 코알라 (APIO17_koala) C++14
29 / 100
134 ms 612 KB
//Never stop trying
/*#pragma GCC target ("avx2")
#pragma GCC optimize ("Ofast")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")*/
#ifndef LOCAL
#include "koala.h"
#endif
#include "bits/stdc++.h"
using namespace std;
#define boost ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)

typedef long long ll;
typedef string str;
typedef double db;
typedef long double ld;
typedef pair<int, int> pi;
#define fi first
#define se second
typedef vector<int> vi;
typedef vector<pi> vpi;
typedef vector<str> vs;
typedef vector<ld> vd;
#define pb push_back
#define eb emplace_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define endl "\n"

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)

const int MOD = 1e9 + 7; //998244353
const ll INF = 1e18;
const int MX = 2e5 + 10;
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //right left down up

template<class T> using V = vector<T>;
template<class T> bool ckmin(T& a, const T& b) { return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }

ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); } // divide a by b rounded up
//constexpr int log2(int x) { return 31 - __builtin_clz(x); } // floor(log2(x))

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
//mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());

ll random(ll a, ll b){
    return a + rng() % (b - a + 1);
}

#ifndef LOCAL  
#define cerr if(false) cerr
#endif
#define dbg(x) cerr << #x << " : " << x << endl; 
#define dbgs(x,y) cerr << #x << " : " << x << " / " << #y << " : " << y << endl;
#define dbgv(v) cerr << #v << " : " << "[ "; for(auto it : v) cerr << it << ' '; cerr << ']' << endl;
#define here() cerr << "here" << endl;

void IO() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
}

int P[100];
int N,W;

#ifdef LOCAL
void playRound(int *B, int *R) {
    int i, j;

    int S = 0;
    for (i=0;i<N;++i) {
        if ( !(B[i] >= 0 && B[i] <= W) ) {
            printf("Invalid query.\n");
            exit(0);
        }
        S += B[i];
    }
    if (S > W) {
        printf("Invalid query.\n");
        exit(0);
    }

    /*numQueries++;
    if (numQueries > maxQueries) {
        printf("Too many queries.\n");
        exit(0);
    }*/

    int cache[2][205];
    int num[2][205];
    char taken[105][205];

    for (i=0;i<205;++i) {
        cache[1][i] = 0;
        num[1][i] = 0;
    }

    for (i=0;i<N;++i) {
        int v = B[i]+1;
        int ii = i&1;
        int o = ii^1;
        for (j=0;j<=W;++j) {
            cache[ii][j] = cache[o][j];
            num[ii][j] = num[o][j];
            taken[i][j] = 0;
        }
        for (j=W;j>=v;--j) {
            int h = cache[o][j-v] + P[i];
            int hn = num[o][j-v] + 1;
            if (h > cache[ii][j] || (h == cache[ii][j] && hn > num[ii][j])) {
                cache[ii][j] = h;
                num[ii][j] = hn;
                taken[i][j] = 1;
            } else {
                taken[i][j] = 0;
            }
        }
    }

    int cur = W;
    for (i=N-1;i>=0;--i) {
        R[i] = taken[i][cur] ? (B[i] + 1) : 0;
        cur -= R[i];
    }
}

#endif

int B[100],R[200];

int minValue(int NN, int WW) {
	N=NN; W=WW;
	/*
	Place only one item in any element:
	-If there is an empty element: it is the MinValue 
	-Else the element in which I placed the item is the MinValue
	*/

	FOR(i,0,N) B[i]=0; 
	B[0]=1;

	playRound(B,R);

	FOR(i,0,N) if(R[i]==0) return i;
	return 0;
}

int maxValue(int NN, int WW) {
	N=NN; W=WW;

	vi chosen;
	for(int i=0; i<100; i+=10){
		FOR(j,0,N) B[j]=0;
		FOR(j,i,i+10) B[j]=10;

		playRound(B,R); 

		FOR(j,i,i+10){
			if(R[j]>10){
				chosen.pb(j);
			}	
		}
	}
	//dbgv(chosen);

	while(sz(chosen)>1){

		FOR(i,0,N) B[i]=0;
		FOR(i,0,min(sz(chosen),10)) B[chosen[i]]=10;

		playRound(B,R);
		vi nw;
		for(auto x: chosen) if(R[x]>10) nw.pb(x);
		FOR(i,10,sz(chosen)) nw.pb(chosen[i]);
		chosen.assign(all(nw));
	}

	return chosen[0];
    return 0;
}

int greaterValue(int NN, int WW) {
	N=NN; W=WW;
    /*
	BS([1,13]) on the number of elements put on both of those
	.if she choose both, make it bigger
	.if she choose no one, make it smaller
	.if she choose only one, that one is the biggest
	!!Choose = put on it more than I put
	*/

	int l=0,r=14;
	while(l<=r){
		int m=(l+r)/2;
		FOR(i,0,N) B[i]=0;
		B[0]=m; B[1]=m; 

		playRound(B,R);
		dbg(m);
		dbgv(R);

		if(R[0]>m && R[1]>m) l=m+1;
		else if(R[0]<=m && R[1]<=m) r=m-1;
		else if(R[0]>m) return 0;
		else return 1;
	}
	assert(0);
    return 0;
}

vi adj[101];

bool vis[101];
vi top;

void dfs(int u){
	vis[u]=true;
	for(auto v: adj[u]) if(!vis[v]) dfs(v);
	top.pb(u);
}

vi solve(vi v){
	if(sz(v)==1) return v;
	
	//dbgv(v);
	vi a,b;

	/*int val=0;
	do{
		val++;
		FOR(i,0,N) B[i]=0;
		for(auto x: v) B[x]=val;

		playRound(B,R);
		a.clear(); b.clear();
		for(auto x: v){
			if(R[x]>1) b.pb(x);
			else if(R[x]==0) a.pb(x);
		}
	}while(a.empty() || b.empty());*/

	int l=1,r=min(N/sz(v),13);
	int prev=8478484;
	while(l<=r){

		int val=(l+r)/2; dbg(val);
		FOR(i,0,N) B[i]=0;
		for(auto x: v) B[x]=val;

		playRound(B,R);
		vi aa,bb;
		for(auto x: v){
			if(R[x]>val) bb.pb(x);
			else aa.pb(x);
		}

		if(abs(sz(aa)-sz(bb))>prev) break;
		prev=abs(sz(aa)-sz(bb));
		dbgv(aa); dbgv(bb);
		a.assign(all(aa));
		b.assign(all(bb));

		if(sz(aa)==sz(bb)) break;	

		if(sz(aa)<sz(bb)) r=val-1;
		else l=val+1;
	}

	vi left=solve(a),right=solve(b);

	for(auto x: right) left.pb(x);
	return left;	
	
}

void allValues(int NN, int WW, int *PP) {
	N=NN; W=WW;

	/*vi v[14];
	FOR(i,0,N){
		int l=0,r=13;
		int ans;
		while(l<=r){
			int m=(l+r)/2;
			FOR(idx,0,N) B[idx]=0;
			B[i]=m; 

			playRound(B,R);

			if(R[i]>m){
				ans=m;
				l=m+1;
			}
			else r=m-1;
		}
		v[ans].pb(i);
	}

	//dbgv(v[12]);

	FOR(i,0,14) FOR(j,i+1,14){
		for(auto u: v[i]) for(auto k: v[j]) adj[u].pb(k);
	}

	FOR(i,1,14){
		vi vec=v[i];
		FOR(j,0,sz(vec)) FOR(k,j+1,sz(vec)){
			int u=vec[j],vv=vec[k];

			FOR(idx,0,N) vis[idx]=0;
			dfs(u); if(vis[vv]) continue;
			FOR(idx,0,N) vis[idx]=0;
			dfs(vv); if(vis[u]) continue;

			FOR(idx,0,N) B[idx]=0;
			B[u]=i; B[vv]=i;
			
			playRound(B,R);

			bool a=(R[u]>i),b=(R[vv]>i);
			if(a&&b) dbgs(u,vv);
			if(b){
				adj[u].pb(vv); 
			}
			else{
				adj[vv].pb(u);
			}
		}
	}

	top.clear();
	FOR(i,0,N) vis[i]=0;
	FOR(i,0,N) if(!vis[i]) dfs(i);
	reverse(all(top));
	//dbgv(top);
	FOR(i,0,N) PP[top[i]]=i+1;
    */

	vi v;
	FOR(i,0,N) v.pb(i);
	vi res=solve(v);
	FOR(i,0,N) PP[res[i]]=i+1;

}

#ifdef LOCAL

int main() {
    boost; IO();

    FOR(i,0,100) P[i]=i+1;

    //P[0]=100; P[99]=1;
    //cout << maxValue(100,100) << endl;

	//P[0]=20; P[1]=10; P[9]=1; P[19]=2;
	//dbgv(P);
    allValues(100,100,P);
    dbgv(P);
    
    return 0;
}

#endif



/* Careful!!!
    .Array bounds
    .Infinite loops
    .Uninitialized variables / empty containers
    .Multisets are shit

   Some insights:
    .Binary search
    .Graph representation
    .Write brute force code
    .Change your approach
*/
# 결과 실행 시간 메모리 Grader output
1 Correct 6 ms 364 KB Output is correct
2 Correct 6 ms 364 KB Output is correct
3 Correct 6 ms 364 KB Output is correct
4 Correct 5 ms 364 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Partially correct 51 ms 364 KB Output is partially correct
2 Partially correct 50 ms 484 KB Output is partially correct
3 Partially correct 50 ms 364 KB Output is partially correct
4 Partially correct 49 ms 364 KB Output is partially correct
# 결과 실행 시간 메모리 Grader output
1 Correct 60 ms 384 KB Output is correct
2 Correct 68 ms 544 KB Output is correct
3 Correct 59 ms 364 KB Output is correct
4 Correct 60 ms 364 KB Output is correct
5 Correct 61 ms 364 KB Output is correct
6 Correct 64 ms 364 KB Output is correct
7 Correct 62 ms 544 KB Output is correct
8 Correct 62 ms 364 KB Output is correct
9 Correct 61 ms 364 KB Output is correct
10 Correct 65 ms 364 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Runtime error 1 ms 492 KB Execution killed with signal 8 (could be triggered by violating memory limits)
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 134 ms 612 KB Output isn't correct
2 Halted 0 ms 0 KB -