<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

326. Perspective
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Breaking news! A Russian billionaire has bought a yet undisclosed NBA team. He's planning to invest huge effort and money into making that team the best. And in fact he's been very specific about the expected result: the first place.

Being his advisor, you need to determine whether it's possible for your team to finish first in its division or not.

More formally, the NBA regular season is organized as follows: all teams play some games, in each game one team wins and one team loses. Teams are grouped into divisions, some games are between the teams in the same division, and some are between the teams in different divisions.

Given the current score and the total number of remaining games for each team of your division, and the number of remaining games between each pair of teams in your division, determine if it's possible for your team to score at least as much wins as any other team in your division.

Input
The first line of input contains N (2 ≤ N ≤ 20) — the number of teams in your division. They are numbered from 1 to N, your team has number 1.

The second line of input contains N integers w1, w2,..., wN, where wi is the total number of games that ith team has won to the moment.

The third line of input contains N integers r1, r2,..., rN, where ri is the total number of remaining games for the ith team (including the games inside the division).

The next N lines contain N integers each. The jth integer in the ith line of those contains aij — the number of games remaining between teams i and j. It is always true that aij=aji and aii=0, for all i ai1 + ai2 +... + aiN ≤ ri.

All the numbers in input are non-negative and don't exceed 10\,000.

Output
On the only line of output, print "YES" (without quotes) if it's possible for the team 1 to score at least as much wins as any other team of its division, and "NO" (without quotes) otherwise.

Example(s)
sample input
sample output
3
1 2 2
1 1 1
0 0 0
0 0 0
0 0 0
YES

sample input
sample output
3
1 2 2
1 1 1
0 0 0
0 0 1
0 1 0
NO

<|response|>
1. Abridged Problem Statement
- There are N teams in a division, numbered 1…N. Your team is team 1.
- Team i currently has w[i] wins and has r[i] games remaining (some against division rivals, some outside).
- For each pair (i,j), a[i][j] is the number of remaining games between team i and team j (so a[i][j]=a[j][i], a[i][i]=0, and ∑ₖa[i][k] ≤ r[i]).
- Each remaining game produces exactly one win for one side.
- Question: Can we assign winners to all remaining games so that team 1's final wins ≥ every other team's final wins?
- Output "YES" if it's possible, otherwise "NO."

2. Key Observations
- Team 1's best‐case total wins is
    W1_max = w[1] + r[1],
  by imagining team 1 wins all its remaining games.
- If any other team i already has w[i] > W1_max, elimination is immediate → answer NO.
- We only need to worry about games among the other teams (2…N).  Games involving team 1 can all be "given" to team 1 in the best scenario.
- We must distribute the wins of the games among teams 2…N so that each of them ends up with at most W1_max wins.
- This is a classic "baseball elimination" setup reducible to a max‐flow feasibility check.

3. Full Solution Approach
a) Compute W1_max = w[1] + r[1].
b) Early exit: if ∃ i ≥ 2 with w[i] > W1_max, print NO.
c) Enumerate all remaining games among teams 2…N.  Let G = list of tuples (i,j,g_ij) with g_ij = a[i][j] > 0 and 2 ≤ i < j ≤ N.  Let totalGames = ∑ g_ij.
d) Build a flow network:
   - Create a source node S and a sink node T.
   - For each game (i,j,g), create a "game node" p.  Add an edge S→p of capacity g, representing these g games must be assigned.
   - From p, add edges p→i and p→j, each of capacity g, meaning each of these g games gives one win to either i or j.
   - For each team node i = 2…N, add an edge i→T of capacity (W1_max − w[i]), limiting how many additional wins team i can collect without surpassing W1_max.
e) Compute max‐flow from S to T (e.g., with Dinic).
   - If max_flow == totalGames, we can assign all inter‐opponent games without any team exceeding W1_max → print YES.
   - Otherwise, print NO.

Complexities:
- Number of game nodes ≤ (N−1 choose 2) = O(N²).
- Edges O(N²).
- Dinic runs in O(E·√V) or better, easily fast for N ≤ 20.

4. C++ Implementation

```cpp
#include <bits/stdc++.h>
// #include <coding_library/graph/maxflow.hpp>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

template<class T>
class MaxFlow {
  private:
    struct Edge {
        T flow, cap;
        int idx, rev, to;
        Edge(int _to, int _rev, T _flow, T _cap, int _idx)
            : to(_to), rev(_rev), flow(_flow), cap(_cap), idx(_idx) {}
    };

    vector<int> dist, po;
    int n;

    bool bfs(int s, int t) {
        fill(dist.begin(), dist.end(), -1);
        fill(po.begin(), po.end(), 0);

        queue<int> q;
        q.push(s);
        dist[s] = 0;

        while(!q.empty()) {
            int u = q.front();
            q.pop();

            for(Edge e: adj[u]) {
                if(dist[e.to] == -1 && e.flow < e.cap) {
                    dist[e.to] = dist[u] + 1;
                    q.push(e.to);
                }
            }
        }
        return dist[t] != -1;
    }

    T dfs(int u, int t, T fl = INF) {
        if(u == t) {
            return fl;
        }

        for(; po[u] < (int)adj[u].size(); po[u]++) {
            auto& e = adj[u][po[u]];
            if(dist[e.to] == dist[u] + 1 && e.flow < e.cap) {
                T f = dfs(e.to, t, min(fl, e.cap - e.flow));
                e.flow += f;
                adj[e.to][e.rev].flow -= f;
                if(f > 0) {
                    return f;
                }
            }
        }

        return 0;
    }

  public:
    const static T INF = numeric_limits<T>::max();

    MaxFlow(int n = 0) { init(n); }

    vector<vector<Edge>> adj;

    void init(int _n) {
        n = _n;
        adj.assign(n + 1, {});
        dist.resize(n + 1);
        po.resize(n + 1);
    }

    void add_edge(int u, int v, T w, int idx = -1) {
        adj[u].push_back(Edge(v, adj[v].size(), 0, w, idx));
        adj[v].push_back(Edge(u, adj[u].size() - 1, 0, 0, -1));
    }

    T flow(int s, int t) {
        assert(s != t);

        T ret = 0, to_add;
        while(bfs(s, t)) {
            while((to_add = dfs(s, t))) {
                ret += to_add;
            }
        }

        return ret;
    }
};

int n;
vector<int> w, r;
vector<vector<int>> cnt;

void read() {
    cin >> n;
    w.resize(n);
    r.resize(n);
    cnt.assign(n, vector<int>(n, 0));

    cin >> w >> r;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            cin >> cnt[i][j];
        }
    }
}

void solve() {
    // Classic baseball-elimination via max flow. Team 1 can finish first only
    // if it wins all of its remaining games, giving it cnt_win_a = w[0]+r[0]
    // wins. Every other game between teams i and j (i,j != 1) must be assigned
    // to one of the two, and no other team i may exceed cnt_win_a wins.
    //
    // Build a flow network: source -> a node per remaining game pair (i,j)
    // with capacity equal to the number of remaining games between them; that
    // pair node -> team i and -> team j (capacity = those games, so the wins
    // can go either way); team i -> sink with capacity cnt_win_a - w[i], the
    // headroom team i has before tying team 1. If any team already has more
    // wins than cnt_win_a the answer is immediately NO.
    //
    // The answer is YES iff every remaining game can be routed, i.e. the max
    // flow equals the total number of games to distribute.

    int cnt_win_a = w[0] + r[0];

    int pairs = n * n;
    MaxFlow<int> mf(n + pairs + 2);
    int s = n + pairs, t = n + pairs + 1;

    for(int i = 1; i < n; i++) {
        if(w[i] > cnt_win_a) {
            cout << "NO" << endl;
            return;
        }

        int max_wins = cnt_win_a - w[i];
        mf.add_edge(i, t, max_wins);
    }

    int need_flow = 0;
    for(int i = 1; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            if(!cnt[i][j]) {
                continue;
            }

            need_flow += cnt[i][j];
            mf.add_edge(s, n + (i * n + j), cnt[i][j]);
            mf.add_edge(n + (i * n + j), i, cnt[i][j]);
            mf.add_edge(n + (i * n + j), j, cnt[i][j]);
        }
    }

    if(mf.flow(s, t) < need_flow) {
        cout << "NO" << endl;
    } else {
        cout << "YES" << endl;
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python Implementation with Detailed Comments
```python
import sys
from collections import deque

class Dinic:
    def __init__(self, n):
        self.n = n
        self.adj = [[] for _ in range(n)]
        self.level = [0]*n
        self.it    = [0]*n

    def add_edge(self, u, v, cap):
        # forward edge: [to, cap, rev-index]
        self.adj[u].append([v, cap, len(self.adj[v])])
        # backward edge: cap=0 initially
        self.adj[v].append([u, 0,   len(self.adj[u]) - 1])

    def bfs(self, s, t):
        for i in range(self.n):
            self.level[i] = -1
        queue = deque([s])
        self.level[s] = 0
        while queue:
            u = queue.popleft()
            for v, cap, rev in self.adj[u]:
                if cap > 0 and self.level[v] < 0:
                    self.level[v] = self.level[u] + 1
                    queue.append(v)
        return self.level[t] >= 0

    def dfs(self, u, t, f):
        if u == t:
            return f
        for i in range(self.it[u], len(self.adj[u])):
            v, cap, rev = self.adj[u][i]
            if cap > 0 and self.level[v] == self.level[u] + 1:
                pushed = self.dfs(v, t, min(f, cap))
                if pushed:
                    # reduce forward
                    self.adj[u][i][1] -= pushed
                    # increase backward
                    self.adj[v][rev][1] += pushed
                    return pushed
            self.it[u] += 1
        return 0

    def max_flow(self, s, t):
        flow = 0
        INF = 10**18
        while self.bfs(s, t):
            self.it = [0]*self.n
            while True:
                pushed = self.dfs(s, t, INF)
                if not pushed:
                    break
                flow += pushed
        return flow

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    N = int(next(it))
    w = [int(next(it)) for _ in range(N)]
    r = [int(next(it)) for _ in range(N)]
    a = [[int(next(it)) for _ in range(N)] for __ in range(N)]

    # 1) Max possible wins for team 1
    W1_max = w[0] + r[0]

    # 2) Immediate elimination
    for i in range(1, N):
        if w[i] > W1_max:
            print("NO")
            return

    # 3) Collect games among opponents 2..N
    games = []
    totalGames = 0
    for i in range(1, N):
        for j in range(i+1, N):
            g = a[i][j]
            if g > 0:
                games.append((i, j, g))
                totalGames += g

    # 4) Build flow network
    M = len(games)
    S = N + M
    T = N + M + 1
    dinic = Dinic(N + M + 2)

    # a) Source -> game nodes
    for idx, (i, j, g) in enumerate(games):
        game_node = N + idx
        dinic.add_edge(S, game_node, g)
        dinic.add_edge(game_node, i,       g)
        dinic.add_edge(game_node, j,       g)

    # b) Team i -> sink
    for i in range(1, N):
        cap = W1_max - w[i]
        dinic.add_edge(i, T, cap)

    # 5) Compute max flow
    flow = dinic.max_flow(S, T)

    # 6) Check if we can assign all games
    print("YES" if flow == totalGames else "NO")

if __name__ == "__main__":
    main()
```

Explanation Highlights:
- We never explicitly model games involving team 1; we assume team 1 wins them all to maximize its tally.
- All remaining inter‐opponent games must be "routed" through the flow from S to T. If every such game can be assigned without any opponent exceeding W1_max, team 1 can finish at least tied for first.
- The flow value must match the total number of those games for feasibility.
