이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
using namespace std;
const int INF = 1e9;
const int BLK = 200;
int n, m;
vector<int> b, p;
vector<vector<int>> v;
int f(int x, int c, int y) {
return x + c * y;
}
void addDoge(int i, int pos, int d, vector<vector<int>> &dist, priority_queue<tuple<int, int, int>> &q) {
if (p[i] < BLK) {
dist[pos][p[i]] = min(dist[pos][p[i]], -d);
q.push(make_tuple(d, pos, p[i]));
} else {
int l = b[i] - p[i], r = b[i] + p[i], x = 1;
while (l >= 0 or r < n) {
if (l >= 0) {
dist[l][0] = min(dist[l][0], -d + x);
q.push(make_tuple(d - x, l, 0));
}
if (r < n) {
dist[r][0] = min(dist[r][0], -d + x);
q.push(make_tuple(d - x, r, 0));
}
l -= p[i];
r += p[i];
x++;
}
}
}
int solve() {
vector<vector<int>> dist(n, vector<int>(BLK, INF));
priority_queue<tuple<int, int, int>> q;
dist[b[0]][p[0]] = 0;
q.push(make_tuple(0, b[0], p[0]));
while (not q.empty()) {
auto [d, pos, power] = q.top();
q.pop();
if (dist[pos][power] != -d) continue;
while (not v[pos].empty()) {
addDoge(v[pos].back(), pos, d, dist, q);
v[pos].pop_back();
}
if (pos + power < n) {
dist[pos + power][power] = min(dist[pos + power][power], -d + 1);
q.push(make_tuple(d - 1, pos + power, power));
}
if (pos - power >= 0) {
dist[pos - power][power] = min(dist[pos - power][power], -d + 1);
q.push(make_tuple(d - 1, pos - power, power));
}
}
int ret = INF;
for (int i = 0; i < BLK; i++) {
ret = min(ret, dist[b[1]][i]);
}
return ret == INF ? -1 : ret;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n >> m;
v.resize(n), b.resize(m), p.resize(m);
cin >> b[0] >> p[0];
for (int i = 1; i < m; i++) {
cin >> b[i] >> p[i];
v[b[i]].push_back(i);
}
cout << solve() << '\n';
return 0;
}
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