有一个n个点的无向图，有m次查询，每次查询给出一些(x_i,y_i)

令dist(x,y)表示x和y点在图中最短距离，dist(x,x)=0,如果x,y不连通则dist(x,y) = inf

每次查询图中有多少个点v与至少一个这次询问给出的(x_i,y_i)满足dist(v,x_i)<=yi

#include <bits/stdc++.h>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

#include <iomanip>

#define ALL(x) (x).begin(), (x).end()

#define rt return

#define dll(x) scanf("%I64d",&x)

#define xll(x) printf("%I64d\n",x)

#define sz(a) int(a.size())

#define all(a) a.begin(), a.end()

#define rep(i,x,n) for(int i=x;i<n;i++)

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define eps 1e-6

#define gg(x) getInt(&x)

#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;

using namespace std;

typedef long long ll;

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

ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

ll powmod(ll a, ll b, ll MOD) {ll ans = ; while (b) {if (b % )ans = ans * a % MOD; a = a * a % MOD; b /= ;} return ans;}

inline void getInt(int* p);

const int maxn = ;

const int inf = 0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

int n;

int dis[maxn][maxn];

bitset<maxn> bs[maxn][maxn];

int m;

int q;

bitset<maxn> res;

std::vector<int> G[maxn];

void bfs(int x)

{

    dis[x][x] = ;

    queue<int> q;

    q.push(x);

    while (!q.empty())

    {

        int u = q.front();

        q.pop();

        for (auto y : G[u])

        {

            if (~dis[x][y])continue;

            dis[x][y] = dis[x][u] + ;

            q.push(y);

        }

    }

}

int main()

{

    // freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);

    // freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);

    memset(dis, -, sizeof(dis));

    gbtb;

    cin >> n >> m >> q;

    int x, y;

    repd(t, , m)

    {

        cin >> x >> y;

        G[x].push_back(y);

        G[y].push_back(x);

    }

    repd(i, , n)

    {

        bfs(i);

        repd(j, , n)

        {

            bs[i][dis[i][j]][j] = ;

        }

        repd(j, , n)

        {

            bs[i][j] |= bs[i][j - ];

        }

    }

    int num;

    repd(t, , q)

    {

        cin >> num;

        res.reset();

        repd(i, , num)

        {

            cin >> x >> y;

            res |= bs[x][y];

        }

        cout << res.count() << endl;

    }

    return ;

}

inline void getInt(int* p) {

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '');

        while ((ch = getchar()) >= '' && ch <= '') {

            *p = *p *  - ch + '';

        }

    }

    else {

        *p = ch - '';

        while ((ch = getchar()) >= '' && ch <= '') {

            *p = *p *  + ch - '';

        }

    }

}