#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
const int N = 500, M = 2e6 + 5, INF = 0x3f3f3f3f;
int h[N], e[M], ne[M], w[M], d[N], cur[N], n;
int idx, S, T, a1, a2, an, b1, b2, bn, res1, res2;
char ch;
void add(int a, int b, int c)
{
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
    e[idx] = a, ne[idx] = h[b], w[idx] = 0, h[b] = idx++;
}
bool bfs()
{
    queue<int> q;
    memset(d, -1, sizeof(d));
    q.push(S), d[S] = 0, cur[S] = h[S];
    while (q.size())
    {
        int t = q.front();
        q.pop();
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (d[j] == -1 && w[i])
            {
                d[j] = d[t] + 1, cur[j] = h[j];
                if (j == T)
                    return true;
                q.push(j);
            }
        }
    }
    return false;
}
int find(int u, int limit)
{
    if (u == T)
        return limit;
    int flow = 0;
    for (int i = cur[u]; ~i && flow < limit; i = ne[i])
    {
        int j = e[i];
        cur[u] = i;
        if (d[j] == d[u] + 1 && w[i])
        {
            int t = find(j, min(limit - flow, w[i]));
            if (!t)
                d[j] = -1;
            w[i] -= t, w[i ^ 1] += t, flow += t;
        }
    }
    return flow;
}
int Dinic()
{
    int r = 0, flow;
    while (bfs())
        while (flow = find(S, INF))
            r += flow;
    return r;
}
int main()
{
    while (cin >> n >> a1 >> a2 >> an >> b1 >> b2 >> bn)
    {
        a1++, a2++, b1++, b2++;
        memset(h, -1, sizeof(h)), idx = S = 0, T = 2 * n + 1;
        add(S, a1, an), add(a1 + n, T, an), add(S, b1, bn), add(b1 + n, T, bn);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
            {
                cin >> ch;
                if (ch == 'O')
                    add(i, j, 2), add(j + n, i + n, 2);
                else if (ch == 'N')
                    add(i, j, INF), add(j + n, i + n, INF);
            }
        add(a2, a2 + n, an), add(b2, b2 + n, bn);
        int res = Dinic();
        cout << "res:" << res << endl;
        if (res == an + bn)
            cout << "Yes\n";
        else
            cout << "No\n";
    }
}

难道我理解错了吗，感觉代码写的很清晰了。思路是把一个来回拆成正向路和反向路，以[1,n]和[n+1,2n]为两个领域，以a2和b2为衔接，同时规定了an,bn的流量限制跑最大流。然而输出的最大流一直是满的...