CodeForces500A New Year Transportation

最新推荐文章于 2019-08-12 15:28:08 发布

转载最新推荐文章于 2019-08-12 15:28:08 发布 · 113 阅读

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原文链接：http://www.cnblogs.com/KingSann/articles/7505808.html

本文介绍了一个基于新年庆祝活动背景的运输系统挑战问题。在一个由n个单元格构成的线性世界中，用户需要通过一系列门户从单元格1移动到目标单元格t。每个门户连接两个特定的单元格，但只能单向通行。文章提供了输入输出示例，并附带了判断是否能成功到达目标单元格的算法代码。

New Year is coming in Line World! In this world, there are n cells numbered by integers from 1 to n, as a 1 × n board. People live in cells. However, it was hard to move between distinct cells, because of the difficulty of escaping the cell. People wanted to meet people who live in other cells.

So, user tncks0121 has made a transportation system to move between these cells, to celebrate the New Year. First, he thought of n - 1 positive integers a₁, a₂, ..., a_n - 1. For every integer i where 1 ≤ i ≤ n - 1the condition 1 ≤ a_i ≤ n - i holds. Next, he made n - 1 portals, numbered by integers from 1 to n - 1. The i-th (1 ≤ i ≤ n - 1) portal connects cell i and cell (i + a_i), and one can travel from cell i to cell (i + a_i) using the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot move from cell (i + a_i) to cell i using the i-th portal. It is easy to see that because of condition 1 ≤ a_i ≤ n - i one can't leave the Line World using portals.

Currently, I am standing at cell 1, and I want to go to cell t. However, I don't know whether it is possible to go there. Please determine whether I can go to cell t by only using the construted transportation system.

Input

The first line contains two space-separated integers n (3 ≤ n ≤ 3 × 10⁴) and t (2 ≤ t ≤ n) — the number of cells, and the index of the cell which I want to go to.

The second line contains n - 1 space-separated integers a₁, a₂, ..., a_n - 1 (1 ≤ a_i ≤ n - i). It is guaranteed, that using the given transportation system, one cannot leave the Line World.

Output

If I can go to cell t using the transportation system, print "YES". Otherwise, print "NO".

Example
Input
8 4
1 2 1 2 1 2 1
Output
YES
Input
8 5
1 2 1 2 1 1 1
Output
NO
Note

In the first sample, the visited cells are: 1, 2, 4; so we can successfully visit the cell 4.

In the second sample, the possible cells to visit are: 1, 2, 4, 6, 7, 8; so we can't visit the cell 5, which we want to visit.

从t开始倒着推。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <cmath>
#include <cctype>
#include <string>
#include <cfloat>
#include <stack>
#include <cassert>

using namespace std;

const int N = 1e5 + 10;

int a[N], vis[N], n, t;

int main() {
    scanf("%d%d", &n, &t);
    for(int i = 1 ; i < n ; i ++) {
        scanf("%d", &a[i]);
    }
    vis[t] = 1;
    for(int i = t - 1 ; i >= 1 ; i --) {
        vis[i] = vis[i + a[i]];
    }
    if(vis[1]) {
        puts("YES");
    } else {
        puts("NO");
    }
}

转载于:https://www.cnblogs.com/KingSann/articles/7505808.html