제출 #137503

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
137503		vince	친구 (IOI14_friend)	C++14	69 / 100	126 ms	6844 KiB

friend

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
#include <utility>
#include <vector>
#include <string>
#include <unordered_map>
#include <map>
#include <queue>
#include <set>
#include <stack>

using namespace std;

#define fi first
#define se second
typedef pair<int,int> ii;

int n;
int A[100003], T[100003], H[100003], col[100003];
bool task2 = 1, task3 = 1, task4 = 1, task5 = 1;

int con[1003][1003];
void create_graph()
{
	for(int i = 1; i < n; i++)
	{
		// printf("%d %d\n", H[i], T[i]);
		if(T[i] == 0) con[i][H[i]] = con[H[i]][i] = 1, col[i] = col[H[i]]^1;
		if(T[i] == 1)
		{
			col[i] = col[H[i]];
			for(int j = 0; j < n; j++)
				con[i][j] = con[j][i] = con[H[i]][j];
		}
		if(T[i] == 2)
		{
			for(int j = 0; j < n; j++)
				con[i][j] = con[j][i] = con[H[i]][j];
			con[i][H[i]] = con[H[i]][i] = 1;
		}
	}
}

//////////////////////////////////////////////////
int mx_bf = 0;
bool in[11];
void bruteforce(int idx, int sum)
{
	if(idx == n)
	{
		bool bol = 1;
		for(int i = 0; i < n; i++)
			for(int j = i+1; j < n; j++)
				if(in[i] && in[j] && con[i][j] == 1)
					bol = 0;
		
		if(bol) mx_bf = max(sum, mx_bf);
	}
	else
	{
		in[idx] = 1;
		bruteforce(idx+1, sum+A[idx]);
		in[idx] = 0;
		bruteforce(idx+1, sum);
	}
}

int solve_bf()
{
	create_graph();
	bruteforce(0, 0);
	return mx_bf;
}
//////////////////////////////////////////////////

int dp[100003][2];
vector<int> chi[100003];
void dfs_tree(int u)
{
	int sum0 = 0, sum1 = 0;
	for(int v : chi[u])
	{
		dfs_tree(v);
		sum0 += dp[v][0];
		sum1 += max(dp[v][0],dp[v][1]);
	}
	dp[u][0] = sum1;
	dp[u][1] = sum0+A[u];
}

int solve_tree()
{
	for(int i = 1; i < n; i++) chi[H[i]].push_back(i);
	dfs_tree(0);
	return max(dp[0][1], dp[0][0]);
}
//////////////////////////////////////////////////

int match[1003], visited[1003];
int bimatch(int u)
{
	if(visited[u]) return 0;
	// printf("MATCHING %d\n", u);

	visited[u] = 1;
	for(int v = 0; v < n; v++)
	{
		if(!con[u][v]) continue;

		if(match[v] == -1)
		{
			// printf("	MATCHED : %d\n", v);
			match[u] = v;
			match[v] = u;
			return 1;
		}

		if(bimatch(match[v]))
		{
			// printf("	MATCHED : %d\n", v);
			match[u] = v;
			match[v] = u;
			return 1;
		}
	}
	return 0;
}

int solve_bipartide()
{
	create_graph();
	memset(visited, 0, sizeof visited);
	memset(match, -1, sizeof match);

	int hit = 0;
	for(int i = 0; i < n; i++)
	{
		if(col[i]) continue;
		memset(visited, 0, sizeof visited);
		hit += bimatch(i);
		// printf("%d\n", hit);
	}
	return n-hit;
}

//////////////////////////////////////////////////

int solve(int task)
{
	int sum = 0;
	int mx = 0;
	for(int i = 0; i < n; i++)
	{
		sum += A[i];
		mx = max(mx, A[i]);
	}

	if(task == 1) return solve_bf();
	if(task == 2) return sum;
	if(task == 3) return mx;
	if(task == 4) return solve_tree();
	if(task == 5) return solve_bipartide();
	return -1;
}



// Find out best sample
int findSample(int N,int confidence[],int host[],int protocol[])
{
	n = N;
	for(int i = 0; i < n; i++)
	{
		A[i] = confidence[i];
		if(A[i] != 1) task5 = 0;
	}

	for(int i = 1; i < n; i++)
	{
		T[i] = protocol[i], H[i] = host[i];
		if(T[i] != 1) task2 = 0;
		if(T[i] != 2) task3 = 0;
		if(T[i] != 0) task4 = 0;
		if(T[i] == 2) task5 = 0;
		// printf("%d\n", T[i]);
	}
			
	if(task5) return solve(5);
	if(task2) return solve(2);
	if(task3) return solve(3);
	if(task4) return solve(4);
	if(n <= 10) return solve(1);

	return -1;
}

• Subtask 1: 11 / 11
• Subtask 2: 8 / 8
• Subtask 3: 8 / 8
• Subtask 4: 19 / 19
• Subtask 5: 23 / 23
• Subtask 6: 0 / 31