답안 #24628

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
24628	2017-06-10T18:58:02 Z	bill_kondo	CEOI16_icc (CEOI16_icc)	C++14	컴파일 오류	0 ms	0 KB

icc

#include "icc.h"
#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> pii;

const int maxn = 1e2 + 10;

int n;
set <pii> e;

int c[maxn];
int d[maxn];

void acha ()
{
	for (int a = 1; a <= n; ++a)
		for (int b = a + 1; b <= n; ++b)
			if (e.find (pii (a, b)) == e.end())
			{
				c[0] = {a};
				d[0] = {b};

				if (query (1, 1, c, d))
				{
					e.insert (pii (a, b));
					setRoad (a, b);
					return;
				}
			}
}

int pointer;
vector <int> adj[maxn];
bool mrk[maxn];

int node, e[maxn], p;

void combina (int v, int l, int r)
{
	if (l == r)
	{
		node = l;
		return;
	}

	int mid = (l + r) >> 1;

	p = 0;

	for (int i = l; i <= mid; ++i)
		e[p++] = d[i];

	if (query (1, p, c, e))
		combina (v, l, mid);
	else
		combina (v, mid + 1, r);
}

void dfs (int v)
{
	mrk[v] = true;

	for (auto u: adj[v])
		if (!mrk[u])
			dfs (u);
}

void aresta ()
{	
	for (int v = 1; v <= n; ++v)
	{
		c[0] = v;

		for (int u = 1; u <= n; ++u)
			mrk[u] = false;

		dfs (v);

		pointer = 0;
		for (int u = 1; u <= n; ++u)
			if (!mrk[u])
				d[pointer++] = u;

		if (query (1, pointer, c, d))
		{
			combina (v, 0, pointer - 1);
			adj[v].push_back (node);
			adj[node].push_back (v);
			setRoad (v, node);
			return;
		}
	}
}

void run (int N)
{
	n = N;

	for (int i = 1; i <= n - 1; ++i)
	{
		if (N <= 15) acha ();
		else if (N <= 50) aresta ();
	}
}

Compilation message

icc.cpp:37:17: error: conflicting declaration 'int e [110]'
 int node, e[maxn], p;
                 ^
icc.cpp:10:11: note: previous declaration as 'std::set<std::pair<int, int> > e'
 set <pii> e;
           ^
icc.cpp: In function 'void combina(int, int, int)':
icc.cpp:52:4: error: no match for 'operator[]' (operand types are 'std::set<std::pair<int, int> >' and 'int')
   e[p++] = d[i];
    ^
icc.cpp:54:23: error: cannot convert 'std::set<std::pair<int, int> >' to 'int*' for argument '4' to 'int query(int, int, int*, int*)'
  if (query (1, p, c, e))
                       ^