scu

题意：N个点。M条边(2 <= N <= 1000 , 0 <= M <= 10^5)，每一个点有个权值W(0 <= W <= 10^5)，现要去除一些点（不能去掉点0），使得结点 0 与结点 N - 1 不连通，求去掉的点的最小权值和。

题目链接：http://cstest.scu.edu.cn/soj/problem.action?id=3254

——>>这是很明显的最小点权割。。

建图方案：

1）将全部点 i 拆成 i 和 i + N。i -> i + N（容量为Wi）

2）原图中的边 i -> j 变成 i + N -> j（容量为无穷大）

3）0 -> 0 + N（由于原图中的边可能有涉及到0 -> x，这时会拆0）

接着，依据最小割最大流定理。求得最小割。。（又换命名法了我。囧。。）

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>

using std::min;
using std::queue;

const int MAXN = 1000 * 2 + 10;
const int MAXM = 2 * (1000 + 100000) + 10;
const int INF = 0x3f3f3f3f;

struct EDGE
{
    int to;
    int cap;
    int flow;
    int nxt;
};

int N, M;
int hed[MAXN], nxt[MAXM], S, T;
int cur[MAXN], h[MAXN];
bool vis[MAXN];
int ecnt;
EDGE edge[MAXM];

void Init()
{
    ecnt = 0;
    memset(hed, -1, sizeof(hed));
}

void AddEdge(int u, int v, int cap)
{
    edge[ecnt].to = v;
    edge[ecnt].cap = cap;
    edge[ecnt].flow = 0;
    edge[ecnt].nxt = hed[u];
    hed[u] = ecnt++;
}

bool Bfs()
{
    queue<int> qu;

    memset(vis, 0, sizeof(vis));
    qu.push(S);
    vis[S] = true;
    h[S] = 0;
    while (!qu.empty())
    {
        int u = qu.front();
        qu.pop();
        for (int e = hed[u]; e != -1; e = edge[e].nxt)
        {
            int v = edge[e].to;
            if (!vis[v] && edge[e].flow < edge[e].cap)
            {
                h[v] = h[u] + 1;
                vis[v] = true;
                qu.push(v);
            }
        }
    }

    return vis[T];
}

int Dfs(int u, int cap)
{
    if (u == T || cap == 0) return cap;

    int flow = 0, subFlow;
    for (int e = cur[u]; e != -1; e = edge[e].nxt)
    {
        cur[u] = e;
        int v = edge[e].to;
        if (h[v] == h[u] + 1 && (subFlow = Dfs(v, min(cap, edge[e].cap - edge[e].flow))) > 0)
        {
            flow += subFlow;
            edge[e].flow += subFlow;
            edge[e ^ 1].flow -= subFlow;
            cap -= subFlow;
            if (cap == 0) break;
        }
    }

    return flow;
}

int Dinic()
{
    int flow = 0;

    while (Bfs())
    {
        memcpy(cur, hed, sizeof(hed));
        flow += Dfs(S, INF);
    }

    return flow;
}

void Read()
{
    int W;
    scanf("%d%d", &N, &M);
    for (int i = 1; i < N; ++i)
    {
        scanf("%d", &W);
        AddEdge(i, i + N, W);
        AddEdge(i + N, i, 0);
    }
    int P, Q;
    for (int i = 0; i < M; ++i)
    {
        scanf("%d%d", &P, &Q);
        AddEdge(P + N, Q, INF);
        AddEdge(Q, P + N, 0);
    }
    AddEdge(0, N, INF);
    AddEdge(N, 0, 0);
    S = 0;
    T = 2 * N - 1;
}

void Solve()
{
    printf("%d
", Dinic());
}

int main()
{
    int T;

    scanf("%d", &T);
    while (T--)
    {
        Init();
        Read();
        Solve();
    }

    return 0;
}