题目链接
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(nsqrt{n}log(nsqrt{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
",d[T]);
return 0;
}