牛客 82E 无向图中的最短距离 (bitset,bfs)

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 <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e3+10;
int n, m, q, vis[N], d[N];
vector<int> g[N];
bitset<N> f[N][N], ans;
queue<int> que;

int main() {
	scanf("%d%d%d", &n, &m, &q);
	REP(i,1,m) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v),g[v].pb(u);
	}
	REP(i,1,n) {
		REP(j,1,n) vis[j]=0,d[j]=n;
		vis[i] = 1, d[i] = 0, que.push(i);
		while (que.size()) {
			int u = que.front(); que.pop();
			for (int v:g[u]) {
				d[v] = min(d[v], d[u]+1);
                if (vis[v]) continue;
				vis[v] = 1;
				que.push(v);
			}
		}
		REP(j,1,n) f[i][d[j]].set(j);
		REP(j,1,n-1) f[i][j]|=f[i][j-1];
	}
	while (q--) {
		int k;
		scanf("%d", &k);
		ans.reset();
		while (k--) {
			int x, y;
			scanf("%d%d", &x, &y);
			ans |= f[x][min(y,n-1)];
		}
		printf("%d
", (int)ans.count());
	}
}