题解

#include <iostream>
#include <cstdio>
#include <queue>
using namespace std;
struct node{
	int val, step;
};
int a[1005];
bool vis[1005];
int n, m, k;
int bfs(){
	queue <node> q;
	q.push({m, 0});
	while(!q.empty()){
		node head = q.front();
		q.pop();
		if(head.val == k) return head.step;
		if(!vis[head.val + a[head.val]] && head.val + a[head.val] <= n){//向上走
			vis[head.val + a[head.val]] = 1;
			q.push({head.val + a[head.val], head.step + 1});
		}if(!head.val - a[head.val] && head.val - a[head.val] > 1){//向下走
			vis[head.val - a[head.val]] = 1;
			q.push({head.val - a[head.val], head.step + 1});
		}
	}return -1;
}
int main(){
	cin >> n >> m >> k;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
	}cout << bfs();

	return 0;
}

时间复杂度：$O(1000)$