이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pb emplace_back
#define AI(i) begin(i), end(i)
template<class T> bool chmin(T &a, T b) { return b < a && (a = b, true); }
template<class T> bool chmax(T &a, T b) { return a < b && (a = b, true); }
#ifdef KEV
#define DE(args...) kout("[ " + string(#args) + " ] = ", args)
void kout() { cerr << endl; }
template<class T, class ...U> void kout(T a, U ...b) { cerr << a << ' ', kout(b...); }
template<class T> void debug(T L, T R) { while (L != R) cerr << *L << " \n"[next(L) == R], ++L; }
#else
#define DE(...) 0
#define debug(...) 0
#endif
const int MAX_N = 300010, inf = 1e9;
int N;
vector<int> edge[MAX_N];
ll bfs_all(int s) {
vector<int> dis(N, inf); dis[s] = 0;
queue<int> q; q.push(s);
ll res = 0;
while (q.size()) {
int x = q.front(); q.pop();
res += dis[x];
for (int u : edge[x]) if (chmin(dis[u], dis[x] + 1))
q.push(u);
}
return res;
}
int DistanceSum(int N, int *X, int *Y) {
::N = N;
if (N > 2000) return -1;
for (int i = 0;i < N;++i)
for (int j = 0;j < N;++j) if (i != j)
if (abs(X[i]-X[j]) + abs(Y[i]-Y[j]) == 1)
edge[i].pb(j);
ll res = 0;
for (int i = 0;i < N;++i)
res += bfs_all(i);
return res / 2 % 1'000'000'000;
}
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