【BZOJ 4070】【APIO 2015】雅加达的摩天楼

http://www.lydsy.com/JudgeOnline/problem.php?id=4070

分块建图。

对每个\(P_i\)分类讨论，\(P_i>\sqrt N\)则直接连边，边数少于\(\sqrt N\)。

对每个横跨长度\(\leq\sqrt N\)的边，建一条“滑轨”，当\(P_i\leq\sqrt N\)时则把这个点送到滑轨上，可以到任何一个位置下来。一共要建\(\sqrt N\)条滑轨。

最后跑最短路就可以了，uoj上死活过不了hack数据，貌似过了的都没有建图？

时间复杂度\(O(n\sqrt n)\)，如果spfa是\(O(m)\)的话。

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int N = 30003;
const int B = 113;
const int Qtot = N * B - 1;

int nxt[N * B * 5], to[N * B * 5], w[N * B * 5];
int point[N * B], cnt = 0, n, m, S[N], P[N];

void ins(int u, int v, int W) {
	nxt[++cnt] = point[u]; to[cnt] = v; w[cnt] = W; point[u] = cnt;
}

bool inq[N * B];
int dist[N * B], tot;
queue <int> qu;

int spfa(int s, int t) {
	memset(dist, 127, sizeof(int) * (tot + 1));
	dist[s] = 0;
	int u, v, di; qu.push(s); inq[s] = true;
	while (!qu.empty()) {
		inq[u = qu.front()] = false; qu.pop();
		for (int i = point[u]; i; i = nxt[i]) {
			v = to[i];
			if ((di = dist[u] + w[i]) < dist[v]) {
				dist[v] = di;
				if (!inq[v]) {
					inq[v] = true;
					qu.push(v);
				}
			}
		}
	}
	return dist[t] == dist[0] ? -1 : dist[t];
}

int main() {
	scanf("%d%d", &n, &m);
	int Si, Pi; tot = B * n;

	int tt = n;
	for (int i = 1; i < B; ++i)
		for (int j = 1; j <= n; ++j)
			ins(++tt, j, 0);

	for (int i = 1; i < B; ++i)
		for (int j = 1; j <= i; ++j) {
			int tmp = j + i, st = i * n;
			while (tmp <= n) {
				ins(st + tmp - i, st + tmp, 1);
				ins(st + tmp, st + tmp - i, 1);
				tmp += i;
			}
		}

	for (int i = 1; i <= m; ++i) {
		scanf("%d%d", &Si, &Pi);
		S[i] = ++Si; P[i] = Pi;
		if (Pi >= B) {
			for (int tmp = Si + Pi, j = 1; tmp <= n; tmp += Pi, ++j)
				ins(Si, tmp, j);
			for (int tmp = Si - Pi, j = 1; tmp >= 1; tmp -= Pi, ++j)
				ins(Si, tmp, j);
		} else
			ins(Si, Pi * n + Si, 0);
	}

	printf("%d\n", spfa(S[1], S[2]));
	return 0;
}