Submission #294185

#	Time	Username	Problem	Language	Result	Execution time	Memory
294185		khangal	Friend (IOI14_friend)	C++14	100 / 100	37 ms	5624 KiB

friend

#include<bits/stdc++.h>
#include "friend.h"
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef long long ll;
typedef double db;
typedef pair<ll,ll> pl;
typedef vector<ll> vl;
typedef vector<vl> vvl;
#define po pop_back
#define pb push_back
#define mk make_pair
#define mt make_tuple
#define lw lower_bound
#define up upper_bound
#define ff first
#define ss second
#define BOOST ios_base::sync_with_stdio(); cin.tie(0); cout.tie(0);
#define MOD 1000000007
#define MAX 1e18
#define MIN -1e18
#define rep(i,a,b) for(ll i=a;i<=b;i++)
#define per(i,a,b) for(ll i=b;i>=a;i--)
#define con continue
#define freopen freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define PI 3.14159265358979323846264338327950288419716939937510582097494459230781640628
#define read(x) scanf("%lld",&x);
#define print(x) printf("%lld ",x);
#define endl '\n';
typedef tree<ll , null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> indexed_multi_set;
typedef tree<ll , null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
// template< typename T>
// using indexed_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
ll n,m,ans,mid,mn,mx,cnt,T,sum,h1,h2,e[1234567],b,c[1234567],d[1<<20],k,i,j,l,r,h,a[1234567],w,x,y,z,t,xx,yy;
bool used[1234567];
pl p[1234567];
vl v[123456],vec,vv1,vv2;
string s1,s;
ll sz[1234567];
// ll tr[5234567],par[1234567],maxx,arr[1234567];
ll dx[4]={-1,1,0,0},dy[4]={0,0,-1,1},c1[123][123];
map<ll,ll> mp;
int findSample(int n,int confidence[],int host[],int protocol[]){
    int ans=0;
    for(int i = n - 1; i > 0; i--){
        if(protocol[i]==0)ans+=confidence[i],confidence[host[i]]=max(0,confidence[host[i]]-confidence[i]);
        if(protocol[i]==1)confidence[host[i]]+=confidence[i];
        if(protocol[i]==2)confidence[host[i]]=max(confidence[host[i]],confidence[i]);
    }
    return ans+confidence[0];
}

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