Submission #299703

#	Time	Username	Problem	Language	Result	Execution time	Memory
299703		kevlee	Friend (IOI14_friend)	C++17	35 / 100	51 ms	65540 KiB

friend

#include "friend.h"
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mod 1000000007
#define h1 7897897897897897
#define h2 7897466719774591
#define b1 98762051
#define b2 98765431
#define inf 1000000000
#define pi 3.1415926535897932384626
#define LMAX 9223372036854775807
#define ll long long
#define fi first
#define se second
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>
#define vp vector<pii>
#define SET(a, b) memset(a, b, sizeof(a))
#define all(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
vi edges[100005];
int dp[100005][2], a[100005], color[100005], one, two, n;
void dfs(int x, int p) {
  dp[x][1] = a[x];
  for (auto y: edges[x]) {
    if (y != p) {
      dfs(y, x);
      dp[x][1] += dp[y][0];
      dp[x][0] += max(dp[y][0], dp[y][1]);
    }
  }
}
void dfs_color(int x, int p, int c) {
  if (color[x]) return;
  color[x] = c;
  if (c == 1) one++;
  else two++;
  for (auto y: edges[x]) {
    if (y != p) {
      dfs_color(y, x, 3-c);
    }
  }
}
int subtask5() {
  int res = 0;
  FOR(i, 0, n-1) {
    if (!color[i]) {
      one = two = 0;
      dfs_color(i, -1, 1);
      res += max(one, two);
    }
  }
  return res;
}
int subtask4() {
  dfs(0, -1);
  return max(dp[0][0], dp[0][1]);
}
int findSample(int N,int confidence[],int host[],int protocol[]){
  n = N;
  FOR(i, 0, n-1) a[i] = confidence[i];
  FOR(i, 1, n-1) {
    if (protocol[i] == 0) {
      edges[i].pb(host[i]);
      edges[host[i]].pb(i);
    } else if (protocol[i] == 1) {
      for (auto y: edges[host[i]]) {
        edges[i].pb(y);
        edges[y].pb(i);
      }
    } else {
      for (auto y: edges[host[i]]) {
        edges[i].pb(y);
        edges[y].pb(i);
      }
      edges[i].pb(host[i]);
      edges[host[i]].pb(i);
    }
  }
  if (*max_element(a, a+n) == 1) {
    return subtask5();
  }
  if (n == 1) return a[0];
  if (protocol[1] == 0) {
    return subtask4();
  }
  if (protocol[1] == 1) {
    int sum = 0;
    FOR(i, 0, n-1) {
      sum += a[i];
    }
    return sum;
  }
  return *max_element(a, a+n);
}

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