Submission #207253

#	Time	Username	Problem	Language	Result	Execution time	Memory
207253		MvC	Pipes (BOI13_pipes)	C++11	65 / 100	357 ms	131076 KiB

pipes

#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define mkp make_pair
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
typedef long double ld;
const ll INF=0x3f3f3f3f3f3f3f3f;
const ll llinf=(1LL<<62);
const int inf=(1<<30);
const int nmax=5e5+50;
const int mod=1e9+7;
using namespace std;
int n,m,i,ex[nmax],ey[nmax],deg[nmax],vl[nmax],vl1[nmax],rs[nmax],viz[nmax],g[nmax],cur,x,y,st;
vector<int>vc,a[nmax];
queue<int>q;
int main()
{
	//freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
	cin>>n>>m;
	if(m>n)rc(0);
	for(i=1;i<=n;i++)cin>>g[i];
	for(i=1;i<=m;i++)
	{
		cin>>ex[i]>>ey[i];
		a[ex[i]].pb(i);
		a[ey[i]].pb(i);
	}
	for(i=1;i<=n;i++)
	{
		deg[i]=(int)a[i].size();
		if(deg[i]==1)q.push(i),viz[i]=1;
	}
	while(!q.empty())
	{
		x=q.front();
		q.pop();
		for(i=0;i<(int)a[x].size();i++)
		{
			y=ex[a[x][i]]^ey[a[x][i]]^x;
			if(!deg[y])continue;
			rs[a[x][i]]=2*(g[x]-vl[x]);
			vl[y]+=g[x]-vl[x];
			vl[x]=g[x];
			deg[y]--;
			deg[x]--;
			if(deg[y]==1)
			{
				viz[y]=1;
				q.push(y);
			}
		}
	}
	for(i=1;i<=n;i++)if(deg[i])x=i;
	st=x;
	while(1)
	{
		for(i=0;i<(int)a[x].size();i++)
		{
			y=ex[a[x][i]]^ey[a[x][i]]^x;
			if(!deg[y])continue;
			vc.pb(a[x][i]);
			x=y;
			break;
		}
		if(st==x)break;
	}
	if((int)vc.size()&1)rc(0);
	for(i=1;i<=n;i++)vl1[i]=vl[i];
	x=st;
	for(i=0;i<(int)vc.size();i++)
	{
		y=ex[vc[i]]^ey[vc[i]]^x;
		vl1[y]+=g[x]-vl1[x];
		vl1[x]=g[x];
		x=y;
	}
	cur=(g[x]-vl1[x])/2;
	x=st;
	for(i=0;i<(int)vc.size();i++)
	{
		y=ex[vc[i]]^ey[vc[i]]^x;
		cur+=g[x]-vl[x];
		vl[y]+=cur;
		vl[x]+=cur;
		rs[vc[i]]=2*cur;
		cur=0;
		x=y;
	}
	for(i=1;i<=m;i++)cout<<rs[i]<<'\n';
	return 0;
}

• Subtask 1: 30 / 30
• Subtask 2: 35 / 70