Submission #1037573

#	Time	Username	Problem	Language	Result	Execution time	Memory
1037573		8pete8	Islands (IOI08_islands)	C++17	6 / 100	474 ms	131072 KiB

islands

#include<iostream>
#include<stack>
#include<map>
#include<vector>
#include<string>
#include<cassert>
#include<unordered_map>
#include <queue>
#include <cstdint>
#include<cstring>
#include<limits.h>
#include<cmath>
#include<set>
#include<algorithm>
#include <iomanip>
#include<numeric>
#include<bitset>
using namespace std;
#define ll long long
#define f first
#define s second
#define pii pair<int,int>
#define ppii pair<int,pii>
#define vi vector<int>
#define pb push_back
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define F(n) for(int i=0;i<n;i++)
#define lb lower_bound
#define ub upper_bound
#define fastio ios::sync_with_stdio(false);cin.tie(NULL);
#pragma GCC optimize ("03,unroll-lopps")
#define int long long
using namespace std;
const int mod=1e9+7,mxn=1e6+5,inf=1e9,minf=-1e9,lg=25;
int n,m,k,x,y,q,z=0;
vector<pii>adj[mxn+10];
int dp[mxn+10][2][2],on[mxn+10],vis[mxn+10],have=0,up[mxn+10];
void dfs(int cur,int p){
    vis[cur]=on[cur]=1;
    for(auto i:adj[cur]){
        if(i.f==p)continue;
        if(vis[i.f]&&on[i.f]){
            x=cur,y=i.f,z=i.s;
            have++;
        }
        if(!vis[i.f])dfs(i.f,cur);
    }
    on[cur]=0;
}
void solve(int cur,int p){
    int sum[2]={0,0};
    vector<int>change[2];
    for(auto i:adj[cur]){
        if(i.f==p)continue;
        if(have&(cur==x||i.f==y))continue;//delete the edge
        up[i.f]=i.s;
        solve(i.f,cur);
        for(int j=0;j<2;j++){
            sum[j]+=dp[i.f][0][j];
            change[j].pb((dp[i.f][1][j])-dp[i.f][0][j]);
        }
    }
    for(int j=0;j<2;j++)sort(all(change[j]));
    //case where theres no extra edge
    if(!change[0].empty())sum[0]+=change[0].back(),change[0].pop_back();
    dp[cur][1][0]=sum[0]+up[cur];
    if(!change[0].empty())sum[0]+=change[0].back(),change[0].pop_back();
    dp[cur][0][0]=sum[0];
    //have extra edge
    if(!have)return;
    if(cur==y||cur==x){
        dp[cur][1][1]=sum[1]+up[cur];
        if(!change[1].empty())sum[1]+=change[1].back(),change[1].pop_back();
        dp[cur][0][1]=sum[1];
    }
    else{
        if(!change[1].empty())sum[1]+=change[1].back(),change[1].pop_back();
        dp[cur][1][1]=sum[1]+up[cur];
        if(!change[1].empty())sum[1]+=change[1].back(),change[1].pop_back();
        dp[cur][0][1]=sum[1];
    }
}
map<pii,int>mp;
int32_t main(){
    fastio
    cin>>n;
    for(int i=1;i<=n;i++){
        int a,b;cin>>a>>b;
        pii x={min(a,i),max(a,i)};
        if(mp.find(x)==mp.end())mp[x]=b;
        else mp[x]=max(mp[x],b);
    }
    for(auto i:mp){
        adj[i.f.f].pb({i.f.s,i.s});
        adj[i.f.s].pb({i.f.f,i.s});
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        if(vis[i])continue;
        have=0,z=0;
        dfs(i,-1);
        if(have>1)assert(0);//we should only have atmost 1 cycle?
        solve(i,-1);
        int mx=0;
        for(int j=0;j<2;j++)for(int k=0;k<2;k++)mx=max(mx,dp[i][j][k]+(k*z));
        ans+=mx;
    }
    cout<<ans<<'\n';
}
/*
7 
3 8 
7 2 
4 2 
1 4 
1 9 
3 4 
2 3
a node cant be in multiple cycle
** there can be atmost 1 cycle for each comp
compute each comp independtly
how to solve a tree?
we can choose atmost 2 edge for each node
because each node can only be passed once
so dp[x][2]->take the parent edge or not
if y-> child of x
dp[x][0]=get 2 max(dp[y][1]) + other dp[y][0]
dp[x][1]=get 1 max(dp[y][1])+ other dp[y][0]

how to solve with 1 cycle?
we can try deleting 1 edge from the cycle to make a tree
so there are 2 cases delete or not delete if delete then do the dp on tree as usual just dont use that edge
if dont delete then we fix that edge so node a-b connected by that edge will only have get to choose 1 extra edge
*/

Compilation message (stderr)

islands.cpp:32:40: warning: bad option '-funroll-lopps' to pragma 'optimize' [-Wpragmas]
   32 | #pragma GCC optimize ("03,unroll-lopps")
      |                                        ^
islands.cpp:39:23: warning: bad option '-funroll-lopps' to attribute 'optimize' [-Wattributes]
   39 | void dfs(int cur,int p){
      |                       ^
islands.cpp:51:25: warning: bad option '-funroll-lopps' to attribute 'optimize' [-Wattributes]
   51 | void solve(int cur,int p){
      |                         ^
islands.cpp:85:14: warning: bad option '-funroll-lopps' to attribute 'optimize' [-Wattributes]
   85 | int32_t main(){
      |              ^

• Subtask 1: 6 / 33
• Subtask 2: 0 / 3
• Subtask 3: 0 / 4
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10