題解

如果一個聯通塊是一個樹的話，方案數就一種，如果這個聯通塊還有別的邊，那選了一條別的邊就會把樹上對應路徑全部取反，所以方案數是\(2^{m - n + 1}\)
如果聯通塊數為\(c\)方案數為\(2^{m - m + c}\)
一個聯通塊有奇數個黑點一定為0

然後就對於每個點判斷是不是割點，是的話看看分成的聯通塊有幾個包含奇數個黑點

然後如果不是割點看看刪掉這個點後聯通塊黑點個數的奇偶性改沒改變

同時要特判聯通塊裡只有一個點的情況

程式碼

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define MAXN 100005
#define mo 974711
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,M,pw[200005];
char s[MAXN];
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,conn,deg[MAXN],rec[MAXN],incodd[MAXN],incconn[MAXN];
int dfn[MAXN],low[MAXN],idx,sta[MAXN],top,siz[MAXN],black,odd;
bool cut[MAXN],vis[MAXN],acs[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void Tarjan(int u,int fa) {
    int son = 0;
    low[u] = dfn[u] = ++idx;
    sta[++top] = u;
    siz[u] = (s[u] == '1');
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(v != fa) {
        if(dfn[v]) low[u] = min(low[u],dfn[v]);
        else {
        Tarjan(v,u);
        if(low[v] >= dfn[u]) {
            ++son;
            if(fa) cut[u] = 1;
            int t = 0;
            while(1) {
            int x = sta[top--];
            t += siz[x];
            if(x == v) break;
            }
            incodd[u] += (t & 1);
            incconn[u]++;
            siz[u] += t;
        }
        low[u] = min(low[v],low[u]);
        }
    }
    }
    if(!fa && son > 1) cut[u] = 1;
    if(cut[u]) {
    if(fa) incconn[u]++;
    incodd[u] += (black - siz[u] & 1);
    if(odd - (black & 1) + incodd[u] == 0) acs[u] = 1;
    }
    else {
    if(odd - (black & 1) + (black - (s[u] == '1') & 1) == 0) acs[u] = 1;
    
    }
}
void dfs(int u) {
    vis[u] = 1;
    if(s[u] == '1') ++black;
    for(int i = head[u] ; i ; i = E[i].next) {
    int v = E[i].to;
    if(!vis[v]) dfs(v);
    }
}
void Init() {
    read(N);read(M);
    memset(head,0,sizeof(head));sumE = 0;
    memset(deg,0,sizeof(deg));
    idx = 0;top = 0;conn = 0;
    memset(dfn,0,sizeof(dfn));memset(low,0,sizeof(low));
    memset(siz,0,sizeof(siz));
    memset(cut,0,sizeof(cut));memset(vis,0,sizeof(vis));
    memset(acs,0,sizeof(acs));
    memset(incodd,0,sizeof(incodd));memset(incconn,0,sizeof(incconn));
    int u,v;
    for(int i = 1 ; i <= M ; ++i) {
    read(u);read(v);
    add(u,v);add(v,u);
    deg[u]++;deg[v]++;
    }
    scanf("%s",s + 1);
    odd = 0;
    for(int i = 1 ; i <= N ; ++i) {
    if(!vis[i]) {
        black = 0;
        ++conn;
        dfs(i);
        rec[i] = black;
        if(black & 1)++odd;
    }
    }
}
void Solve() {
    if(odd >= 2) {
    for(int i = 1 ; i <= N + 1 ; ++i) {out(0);space;}
    return;
    }
    for(int i = 1 ; i <= N ; ++i) {
    if(!dfn[i]) {
        top = 0;idx = 0;
        black = rec[i];
        Tarjan(i,0);
        if(!deg[i]) cut[i] = 1;
    }
    }
    if(odd == 0) {
    out(pw[M - N + conn]);space;
    }
    else {
    out(0);space;
    }
    for(int i = 1 ; i <= N ; ++i) {
    if(!acs[i]) out(0);
    else if(cut[i]) {
        out(pw[(M - deg[i]) - (N - 1) - 1 + conn + incconn[i]]);
    }
    else {
        out(pw[(M - deg[i]) - (N - 1) + conn]);
    }
    space;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    pw[0] = 1;
    for(int i = 1 ; i <= 200000; ++i) {
    pw[i] = mul(pw[i - 1],2);
    }
    int T;
    read(T);
    while(T--) {Init();Solve();enter;}
}