题解

事实上是个分块暴力

就是跳跃长度大于\(\sqrt{n}\)的狗最多有\(\sqrt{n}\)个位置
剩下跳跃长度小于\(\sqrt{n}\)的暴力记录

也就是两个\(dis\)数组
\(dis[0][i][j]\)表示第\(i\)个位置有跳跃长度为\(j\)的狗
\(dis[1][i][j]\)表示第\(i\)只狗在自己的第\(j\)个位置上

然后遇到一个新的位置把这个位置上所有原有的狗全部扔进队列
之后BFS就行

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define eps 1e-8
#define mo 974711
#define MAXN 30005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
int dis[2][MAXN][205];
bool vis[MAXN][205],rec[MAXN];
vector<int> v[MAXN];
vector<int> dog[MAXN];
int M,N;
int B[MAXN],P[MAXN],st[MAXN];struct node {int a,b,c;
};
queue<node> Q;
void Process(int t,int d) {if(!rec[t]) {int s = dog[t].size();for(int i = 0 ; i < s ; ++i) {int h = dog[t][i];if(dis[1][h][st[h]] == -1) {dis[1][h][st[h]] = d;Q.push((node){1,h,st[h]});}}for(int i = 1 ; i < 200 ; ++i) {if(vis[t][i]) {if(dis[0][t][i] == -1) {dis[0][t][i] = d;Q.push((node){0,t,i});}}}rec[t] = 1;}
}
void BFS() {memset(dis,-1,sizeof(dis));if(P[0] < 200) {dis[0][B[0]][P[0]] = 0;Q.push((node){0,B[0],P[0]});}else {dis[1][0][st[0]] = 0;Q.push((node){1,0,st[0]});}Process(B[0],0);while(!Q.empty()) {node u = Q.front();Q.pop();int t;int d = dis[u.a][u.b][u.c];if(u.a == 0) {if(u.b + u.c < N) {if(dis[0][u.b + u.c][u.c] == -1) {dis[0][u.b + u.c][u.c] = d + 1;Q.push((node){0,u.b + u.c,u.c});Process(u.b + u.c,d + 1);}}if(u.b - u.c >= 0) {if(dis[0][u.b - u.c][u.c] == -1) {dis[0][u.b - u.c][u.c] = d + 1;Q.push((node){0,u.b - u.c,u.c});Process(u.b - u.c,d + 1);}}}else {if(u.c < v[u.b].size() - 1) {if(dis[1][u.b][u.c + 1] == -1) {dis[1][u.b][u.c + 1] = d + 1;Q.push((node){1,u.b,u.c + 1});Process(v[u.b][u.c + 1],d + 1);}}if(u.c > 0) {if(dis[1][u.b][u.c - 1] == -1) {dis[1][u.b][u.c - 1] = d + 1;Q.push((node){1,u.b,u.c - 1});Process(v[u.b][u.c - 1],d + 1);}}}}
}
void Init() {read(N);read(M);for(int i = 0 ; i < M ; ++i) {read(B[i]);read(P[i]);if(P[i] >= 200) {int k = B[i];dog[B[i]].pb(i);while(k >= P[i]) k -= P[i];while(k < N) {v[i].pb(k);if(k == B[i]) st[i] = v[i].size() - 1;k += P[i];}}else vis[B[i]][P[i]] = 1;}
}
void Solve() {BFS();int ans;if(P[1] >= 200) ans = dis[1][1][st[1]];else ans = dis[0][B[1]][P[1]];out(ans);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifInit();Solve();
}