2020牛客暑期多校训练营（第一场) I 1 or 2

给一个图，要求删一些边，使每个点的度数为给定的$d_i(1\le d_i\le 2)$。

每个点拆成$i$和$i^{\prime }$，源点$S$和$i$连一条权值为$d_i$的边，$i^{\prime }$和汇点$T$连一条权值为$d_i$的边；$m$条边，$u$和$v^{\prime }$、$u‘$和$v$建边。
然后只要判断最大流与$\sum d_i$是否相等即可。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include 

#define mem(ss) memset(ss,0,sizeof(ss))
#define rep(d, s, t) for(int d=s;d<=t;d++)
#define rev(d, s, t) for(int d=s;d>=t;d--)
#define inf 0x3f3f3f3f
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef double db;
typedef std::pair<int, int> pii;
typedef std::pair<ll, ll> pll;
typedef std::pair<double, double> pdd;
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 8e5 + 10;
#define io_opt ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;

ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }

struct mxfl {
    int n = 0;
    int m = 0;
    int s, t;
    struct Edge {
        int from, to, cap, flow;
    };
    vector<Edge> edges;
    vector<int> G[N];

    void init(int _n, int _s, int _t) {
        n = _n;
        s = _s;
        t = _t;
        edges.clear();
        for (int i = 1; i <= n; i++)G[i].clear();
    }

    void add_edge(int from, int to, int cap) {
        edges.push_back(Edge{from, to, cap, 0});
        edges.push_back(Edge{to, from, 0, 0});
        m = (int) edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool vis[N];
    int d[N], cur[N];

    bool bfs() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        d[s] = 0;
        vis[s] = 1;
        while (!Q.empty()) {
            int x = Q.front();
            Q.pop();
            for (int i = 0; i < (int) G[x].size(); i++) {
                Edge &e = edges[G[x][i]];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int dfs(int x, int a) {
        if (x == t || a == 0)return a;
        int flow = 0, f;
        for (int &i = cur[x]; i < (int) G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[G[x][i] ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0)break;
            }
        }
        return flow;
    }

    int maxflow() {
        int flow = 0;
        while (bfs()) {
            memset(cur, 0, sizeof(cur));
            flow += dfs(s, inf);
        }
        return flow;
    }
} MF;

int n, m;
int d[N];

int main() {
    while (scanf("%d%d", &n, &m) == 2) {
        int S = 2 * n + 1, T = 2 * n + 2;
        MF.init(2 * n + 2, S, T);
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &d[i]);
            MF.add_edge(S, i, d[i]);
            MF.add_edge(n + i, T, d[i]);
            sum += d[i];
        }
        for (int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            MF.add_edge(u, v + n, 1);
            MF.add_edge(v, u + n, 1);
        }
        if (MF.maxflow() != sum)puts("No");
        else puts("Yes");
    }
    return 0;
}