BZOJ4070 [Apio2015]雅加达的摩天楼 【分块 + 最短路】

题目链接

BZOJ4070

题解

考虑暴力建图，将每个\(B_i\)向其能到的点连边，复杂度\(O(\sum \frac{n}{p_i})\)，当\(p\)比较小时不适用

考虑优化建图，每个\(doge\)能移动的点实际上是一组模\(p\)同余的点，那么只要对每个\(p\)建\(n\)个点，然后内部距离为\(p\)的点连边，然后每个点向原来的点连边，如果某个点有步长为\(p\)的\(doge\)，则原点向该点连边，这样子每一层点数和边数都是\(O(n)\)的，复杂度是\(O(pn)\)，当\(p\)较小时使用

如此可以得出最终算法，分块处理

对于\(p\)较小的点优化建图，\(p\)较大的点暴力建图

跑最短路即可

如果是\(dijsktra\)，复杂度\(O(n\sqrt{n}log(n\sqrt{n}))\)

但据说这里\(spfa\)快？

#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 4000005,maxm = 15000005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 1;
struct EDGE{int to,nxt,w;}ed[maxm];
inline void build(int u,int v,int w){
	ed[++ne] = (EDGE){v,h[u],w}; h[u] = ne;
}
int n,m,B,d[maxn],inq[maxn],S,T;
queue<int> q;
void spfa(){
	int E = B * n + n;
	for (int i = 1; i <= E; i++) d[i] = INF;
	d[S] = 0; q.push(S); int u;
	while (!q.empty()){
		u = q.front(); q.pop(); inq[u] = false;
		Redge(u) if (d[to = ed[k].to] > d[u] + ed[k].w){
			d[to] = d[u] + ed[k].w;
			if (!inq[to]) q.push(to),inq[to] = true;
		}
	}
}
int main(){
	n = read(); m = read(); B = min((int)sqrt(n),100);
	for (int i = 1; i <= B; i++){
		for (int j = 1; j <= i; j++){
			for (int u = j; u + i <= n; u += i){
				build(i * n + u,i * n + u + i,1);
				build(i * n + u + i,i * n + u,1);
			}
		}
		for (int j = 1; j <= n; j++)
			build(i * n + j,j,0);
	}
	int x,p;
	for (int t = 1; t <= m; t++){
		x = read() + 1; p = read();
		if (t == 1) S = x;
		if (t == 2) T = x;
		if (p > B){
			for (int i = p,j = 1; x - i > 0; i += p,j++)
				build(x,x - i,j);
			for (int i = p,j = 1; x + i <= n; i += p,j++)
				build(x,x + i,j);
		}
		else build(x,p * n + x,0);
	}
	spfa();
	if (d[T] == INF) puts("-1");
	else printf("%d\n",d[T]);
	return 0;
}