# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
1162454 | 8pete8 | Brought Down the Grading Server? (CEOI23_balance) | C++20 | 62 ms | 29764 KiB |
#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<complex>
#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
#define double long double
using namespace std;
using cd = complex<double>;
double const PI=acos(-1);
const int mod=1e9+7,mxn=5e5+5,inf=1e17,minf=-1e9,lg=15,base=311;
//#undef int
int k,m,n,q,d,S,t;
void setIO(string name){
ios_base::sync_with_stdio(0); cin.tie(0);
freopen((name+".in").c_str(),"r",stdin);
freopen((name+".out").c_str(),"w",stdout);
}
vector<int>row[mxn+10],have[mxn+10];
int can[mxn+10],cnt[mxn+10],yes[mxn+10];
pii from[mxn+10];
int imbalance[mxn+10];
int tol=0;
int get(int x){
if(abs(imbalance[x])==cnt[x])return 0;
return abs(imbalance[x]);
}
int32_t main(){
cin>>n>>S>>t;
for(int i=1;i<=n;i++){
row[i].resize(S);
for(auto &j:row[i])cin>>j,cnt[j]++;
imbalance[row[i][0]]++;
imbalance[row[i][1]]--;
}
for(int i=1;i<=t;i++)tol+=get(i);
for(int i=1;i<=n;i++){
int x=get(imbalance[row[i][0]])+get(imbalance[row[i][1]]);
imbalance[row[i][0]]-=2;
imbalance[row[i][1]]+=2;
int y=get(imbalance[row[i][0]])+get(imbalance[row[i][1]]);
if(y<x)swap(row[i][0],row[i][1]),tol=tol-x+y;
else{
imbalance[row[i][0]]+=2;
imbalance[row[i][1]]-=2;
}
}
for(int i=1;i<=n;i++)cout<<row[i][0]<<" "<<row[i][1]<<'\n';
}
/*
knapsack for s==2?
for general case can we split into 2 valid groups (by knapsack)
then d&c down to get the answer??
how to get configuration with knapsack??
case like:
2 3
2 3
1 2
1 4
1 4
1 2
we cant take 1 and 4 to be on the same side?
can we just greedy swap??
*/
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