【Good Bye 2014A】【水题 dfs】New Year Transportation 可否传送到t点
2016-03-21 09:06
495 查看
New Year Transportation
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 - 1positive
integers a1, a2, ..., an - 1.
For every integer i where 1 ≤ i ≤ n - 1 the
condition 1 ≤ ai ≤ 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 + ai),
and one can travel from cell i to cell(i + ai) using
the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot
move from cell (i + ai) to
celli using the i-th
portal. It is easy to see that because of condition 1 ≤ ai ≤ 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, ..., an - 1 (1 ≤ ai ≤ 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".
Examples
input
output
input
output
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.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 3e4+10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n, t, x;
vector<int>a
;
bool vis
;
void dfs(int x)
{
vis[x] = 1;
for (int i = a[x].size() - 1; ~i; --i)
{
int y = a[x][i];
if (!vis[y])dfs(y);
}
}
int main()
{
while (~scanf("%d%d", &n, &t))
{
for (int i = 1; i < n; ++i)
{
scanf("%d", &x);
a[i].push_back(i + x);
}
dfs(1);
puts(vis[t] ? "YES" : "NO");
}
return 0;
}
/*
【题意】
有n(3e4)个点,排成一排。
对于点i,可以传送到i+a[i]。
问你我们可否传到t点
【类型】
bfs or dfs
【分析】
扫一下就好啦
【时间复杂度&&优化】
O(n)
*/
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 - 1positive
integers a1, a2, ..., an - 1.
For every integer i where 1 ≤ i ≤ n - 1 the
condition 1 ≤ ai ≤ 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 + ai),
and one can travel from cell i to cell(i + ai) using
the i-th portal. Unfortunately, one cannot use the portal backwards, which means one cannot
move from cell (i + ai) to
celli using the i-th
portal. It is easy to see that because of condition 1 ≤ ai ≤ 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, ..., an - 1 (1 ≤ ai ≤ 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".
Examples
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.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 3e4+10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n, t, x;
vector<int>a
;
bool vis
;
void dfs(int x)
{
vis[x] = 1;
for (int i = a[x].size() - 1; ~i; --i)
{
int y = a[x][i];
if (!vis[y])dfs(y);
}
}
int main()
{
while (~scanf("%d%d", &n, &t))
{
for (int i = 1; i < n; ++i)
{
scanf("%d", &x);
a[i].push_back(i + x);
}
dfs(1);
puts(vis[t] ? "YES" : "NO");
}
return 0;
}
/*
【题意】
有n(3e4)个点,排成一排。
对于点i,可以传送到i+a[i]。
问你我们可否传到t点
【类型】
bfs or dfs
【分析】
扫一下就好啦
【时间复杂度&&优化】
O(n)
*/
相关文章推荐
- 【HDU 5366】The mook jong 详解
- HDU 5240 Exam (好水的题)
- 北大—1006——Biorhythms
- 时间计算(heaven.pas/cpp)
- 工作依赖(job.cpp/pas)
- 题目 英雄 (BFS)
- kmp 学习 hihocoder #1015
- HDU 1096 A+B for Input-Output Practice (VIII)
- HUST-1601 - Shepherd 暴力
- P1478
- P1035
- P1008 难度2.7
- 2020 绝对值排序
- 2021 发工资咯
- 2022 海选女主角
- 2024 C语言合法标识符
- 2025 查找最大元素
- 2026 首字母变大写
- 2027 统计元音
- 2028 Lowest Common Multiple Plus