1 or 2

链接：https://ac.nowcoder.com/acm/contest/5666/I
来源：牛客网

题目描述
Bobo has a graph with $n$ vertices and $m$ edges where the $i$-th edge is between the vertices $a_i$ and $b_i$. Find out whether is possible for him to choose some of the edges such that the $i$-th vertex is incident with exactly $d_i$ edges.
输入描述:
The input consists of several test cases terminated by end-of-file.
The first line of each test case contains two integers $n$ and $m$.
The second line contains $n$ integers $d_1,d_2,\dots ,d_n$.
The $i$-th of the following $m$ lines contains two integers $a_i$ and $b_i$.

• $1\le n\le 50$
• $1\le m\le 100$
• $1\le d_i\le 2$
• $1\le a_i,b_i\le n$
• $a_i\ne b_i$
• $\{a_i,b_i\}\ne \{a_j,b_j\}$ for $i\ne j$
• The number of tests does not exceed $100$.

输出描述:
For each test case, print “Yes” without quotes if it is possible. Otherwise, print “No” without quotes.
示例1

输入
2 1
1 1
1 2
2 1
2 2
1 2
3 2
1 1 2
1 3
2 3
输出
Yes
No
Yes

对于某条边$e_{x,y}$，将边拆成两个点$e_1$，$e_2$，在$e_1$，$e_2$之间连边，在$x$，$e_1$之间连边，$e_2$，$y$之间连边。如果$d[x]=2$，将$x$拆出一个点$x^{\prime }$，在$x^{\prime }$，$e_1$之间连边，点$y$同理。
样例3

3 2
1 1 2
1 3
2 3

建图如下。

对所建图求一般图最大匹配，如果存在完美匹配则输出"Yes"，否则输出"No"。

#include

#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define sc(a) scahf("%c",&a);
#define ss(a) scanf("%s",a)
#define pi(a) printf("%d\n",a)
#define pl(a) printf("%lld\n",a)
#define pc(a) putchar(a)
#define ms(a) memset(a,0,sizeof(a))
#define repi(i, a, b) for(register int i=a;i<=b;++i)
#define repd(i, a, b) for(register int i=a;i>=b;--i)
#define reps(s) for(register int i=head[s];i;i=Next[i])
#define ll long long
#define ull unsigned long long
#define vi vector
#define pii pair
#define mii unordered_map
#define msi unordered_map
#define lowbit(x) ((x)&(-(x)))
#define ce(i, r) i==r?'\n':' '
#define pb push_back
#define fi first
#define se second
#define INF 0x3f3f3f3f
#define pr(x) cout<<#x<<": "<
using namespace std;

inline int qr() {
    int f = 0, fu = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-')fu = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        f = (f << 3) + (f << 1) + c - 48;
        c = getchar();
    }
    return f * fu;
}

const int N = 305;

struct BA {
    int n;
    bool G[N][N];
    int Match[N];
    bool InQueue[N], InPath[N], InBlossom[N];
    int Head, Tail;
    int Queue[N];
    int Start, Finish;
    int NewBase;
    int Father[N], Base[N];
    int Count;

    inline void add(int x, int y) {
        G[x][y] = G[y][x] = true;
    }

    inline void init(int _n) {
        n = _n, ms(G);
    }

    inline void Push(int u) {
        Queue[Tail] = u;
        Tail++;
        InQueue[u] = true;
    }

    inline int Pop() {
        int res = Queue[Head];
        Head++;
        return res;
    }

    inline int FindCommonAncestor(int u, int v) {
        ms(InPath);
        while (true) {
            u = Base[u];
            InPath[u] = true;
            if (u == Start) break;
            u = Father[Match[u]];
        }
        while (true) {
            v = Base[v];
            if (InPath[v])break;
            v = Father[Match[v]];
        }
        return v;
    }

    inline void ResetTrace(int u) {
        int v;
        while (Base[u] != NewBase) {
            v = Match[u];
            InBlossom[Base[u]] = InBlossom[Base[v]] = true;
            u = Father[v];
            if (Base[u] != NewBase) Father[u] = v;
        }
    }

    inline void BloosomContract(int u, int v) {
        NewBase = FindCommonAncestor(u, v);
        ms(InBlossom);
        ResetTrace(u);
        ResetTrace(v);
        if (Base[u] != NewBase) Father[u] = v;
        if (Base[v] != NewBase) Father[v] = u;
        repi(tu, 1, n)if (InBlossom[Base[tu]]) {
                Base[tu] = NewBase;
                if (!InQueue[tu]) Push(tu);
            }
    }

    inline void FindAugmentingPath() {
        ms(InQueue), ms(Father);
        repi(i, 1, n) Base[i] = i;
        Head = Tail = 1;
        Push(Start);
        Finish = 0;
        while (Head < Tail) {
            int u = Pop();
            repi(v, 1, n)if (G[u][v] && (Base[u] != Base[v]) && (Match[u] != v)) {
                    if ((v == Start) || ((Match[v] > 0) && Father[Match[v]] > 0))
                        BloosomContract(u, v);
                    else if (Father[v] == 0) {
                        Father[v] = u;
                        if (Match[v] > 0)
                            Push(Match[v]);
                        else {
                            Finish = v;
                            return;
                        }
                    }
                }
        }
    }

    inline void AugmentPath() {
        int u, v, w;
        u = Finish;
        while (u > 0) {
            v = Father[u];
            w = Match[v];
            Match[v] = u;
            Match[u] = v;
            u = w;
        }
    }

    inline void Edmonds() {
        ms(Match);
        repi(u, 1, n)if (Match[u] == 0) {
                Start = u;
                FindAugmentingPath();
                if (Finish > 0) AugmentPath();
            }
    }

    inline void PrintMatch() {
        Count = 0;
        repi(u, 1, n) if (Match[u] > 0) Count++;
        printf("%d\n", Count);
        repi(u, 1, n) if (u < Match[u])
                printf("%d %d\n", u, Match[u]);
    }
} B;

int n, m;
int d[55];

int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        B.init(n * 2 + m * 2);
        int tot = n * 2, num = m * 2;
        repi(i, 1, n)d[i] = qr(), num += d[i];
        repi(i, 1, m) {
            int x = qr(), y = qr();
            B.add(tot + 1, tot + 2);
            B.add(x, tot + 1), B.add(tot + 2, y);
            if (d[x] == 2)B.add(x + n, tot + 1);
            if (d[y] == 2)B.add(tot + 2, y + n);
            tot += 2;
        }
        B.Edmonds();
        int cnt = 0;
        repi(i, 1, n * 2 + m * 2)if (B.Match[i])cnt++;
        puts(cnt == num ? "Yes" : "No");
    }
    return 0;
}