Description

给出一个nn个点mm条边的图，对于每个点有一个流量sis_i表示它需要恰好流进或流出（正/负）多少流量

问是否存在一种流量方案使得每个点都满足流进或流出sis_i的流量

如果存在则输出方案

Constraints

n,m≤2∗105n,m \leq 2 * 10^5

Solution

考虑贪心，从这个图中提取出一棵树出来，在树上考虑流量问题

那么从下至上依次贪心满足流量，最后判断1号节点流量是否平衡即可

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define x1 X1 
#define x2 X2
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) {return a < b ? a = b, 1 : 0;}
template <typename T> inline int Chkmin (T &a, T b) {return a > b ? a = b, 1 : 0;}
template <typename T> T read()
{
	T fl = 1, sum = 0; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
	return sum * fl;
}

const int Maxn = 2 * 1e5 + 100;

int N, M, e, Begin[Maxn], To[Maxn << 1], Next[Maxn << 1], W[Maxn << 1];
int A[Maxn], Vis[Maxn], Ans[Maxn];

inline void add_edge (int x, int y)
{
	To[++e] = y;
	Next[e] = Begin[x];
	Begin[x] = e;
}

inline void dfs (int x)
{
	Vis[x] = 1;
	for (int i = Begin[x]; i; i = Next[i])
	{
		int y = To[i];
		if (Vis[y]) continue;
		dfs(y);
		Ans[(i + 1) / 2] = A[y] * ((i % 2) ? 1 : -1);
		A[x] += A[y];
	}
}

int main()
{
#ifdef hk_cnyali
	freopen("F.in", "r", stdin);
	freopen("F.out", "w", stdout);
#endif
	N = read<int>();
	for (int i = 1; i <= N; ++i) A[i] = read<int>();
	M = read<int>();
	for (int i = 1; i <= M; ++i)
	{
		int x = read<int>(), y = read<int>();
		add_edge (x, y);
		add_edge (y, x);
	}
	dfs(1);
	if (A[1]) 
	{
		puts("Impossible");
		return 0;
	}
	puts("Possible");
	for (int i = 1; i <= M; ++i) cout<<Ans[i]<<endl;
	return 0;
}