思路：拆點，每個點拆成兩個，限流的不用多說了吧。然後，關鍵在於某點A和B有邊，但是實際結果只能二選一，怎麼辦呢？弄一個輔助點C加邊方式(A’,C,1,-SC[A][B]), (B',C,1,-SC[B][A]), (C,C', 1, 0), (C',T, INF, 0); //(U,V,CAP,COST)

程式碼：

#include <stdio.h>
#include <string.h>
#include <vector>
#include <string>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <set>
#include <iostream>
#include <map>
#include <algorithm>
const double PI = acos(-1.0);
const double EPS = 1e-9;
using namespace std;


const int MAXN = 50000;
const int MAXM = 500000;
const int INF = 0x3f3f3f3f;
struct Edge
{
	int to, next, cap, flow, cost;
}edge[MAXM];
int head[MAXN], tol;
int pre[MAXN], dis[MAXN];
bool vis[MAXN];
int N;//節點總個數，節點編號從0~N-1
void init()
{
	tol = 0;
	memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)
{
	edge[tol].to = v;
	edge[tol].cap = cap;
	edge[tol].cost = cost;
	edge[tol].flow = 0;
	edge[tol].next = head[u];
	head[u] = tol++;
	edge[tol].to = u;
	edge[tol].cap = 0;
	edge[tol].cost = -cost;
	edge[tol].flow = 0;
	edge[tol].next = head[v];
	head[v] = tol++;
}
bool spfa(int s, int t)
{
	queue<int>q;
	for (int i = 0; i < N; i++)
	{
		dis[i] = INF;
		vis[i] = false;
		pre[i] = -1;
	}
	dis[s] = 0;
	vis[s] = true;
	q.push(s);
	while (!q.empty())
	{
		int u = q.front();
		q.pop();
		vis[u] = false;
		for (int i = head[u]; i != -1; i = edge[i].next)
		{
			int v = edge[i].to; 
			if (edge[i].cap > edge[i].flow &&
				dis[v] > dis[u] + edge[i].cost)
			{
				dis[v] = dis[u] + edge[i].cost;
				pre[v] = i;
				if (!vis[v])
				{
					vis[v] = true;
					q.push(v);
				}
			}
		}
	}
	if (pre[t] == -1)return false;
	else return true;
}
//返回的是最大流，cost存的是最小費用
int minCostMaxflow(int s, int t, int &cost)
{
	int flow = 0;
	cost = 0;
	while (spfa(s, t))
	{
		int Min = INF;
		for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
		{
			if (Min > edge[i].cap - edge[i].flow)
				Min = edge[i].cap - edge[i].flow;
		}
		for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
		{
			edge[i].flow += Min;
			edge[i ^ 1].flow -= Min;
			cost += edge[i].cost * Min;
		}
		flow += Min;
	}
	return flow;
}

bool flag[105][105];
int sc[105][105];
int n, st, ed;

void add(int i, int j)
{
	addedge(i+n, N, 1, -sc[i][j]);
	addedge(j+n, N, 1, -sc[j][i]);
	N++;
	addedge(N - 1, N, 1, 0);
	addedge(N, ed, INF, 0);
	N++;
}

int main()
{
	int a[105];
	while (scanf("%d", &n) != EOF)
	{
		ed = 2 * n + 1;
		N = 2 * n + 2;
		memset(flag, 0, sizeof(flag));
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		int m;
		scanf("%d", &m);
		for (int i = 1; i <= m; i++)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			a[x]--;
			flag[x][y] = true;
		}

		int sum = 0;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				scanf("%d", &sc[i][j]);
				if (flag[i][j])
				{
					sum += sc[i][j];
					sc[i][j] = -1;
				}
				if (flag[j][i])
					sc[i][j] = -1;
			}
		}

		init();
		for (int i = 1; i <= n; i++)
		{
			addedge(st, i, INF, 0);
			addedge(i, i + n, a[i], 0);
			for (int j = i+1; j <= n; j++)
			{
				if (sc[i][j] != -1)
				{
					add(i, j);
				}
			}
		}

		int ans = 0;
		minCostMaxflow(st, ed, ans);
		cout << sum-ans << endl;
	}
	return 0;
}