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(xi,yi)
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(xi,yi)满足dist(v,xi)<=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 '\n' #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\n", (int)ans.count()); } }