CodeForces 689B Mike and Shortcuts (bfs or 最短路)
Mike and Shortcuts
题目链接:
http://acm.hust.edu.cn/vjudge/contest/121333#problem/F
Description
Recently, Mike was very busy with studying for exams and contests. Now he is going to chill a bit by doing some sight seeing in the city.
City consists of n intersections numbered from 1 to n. Mike starts walking from his house located at the intersection number 1 and goes along some sequence of intersections. Walking from intersection number i to intersection j requires |i - j| units of energy. The total energy spent by Mike to visit a sequence of intersections p1 = 1, p2, ..., pk is equal to units of energy.
Of course, walking would be boring if there were no shortcuts. A shortcut is a special path that allows Mike walking from one intersection to another requiring only 1 unit of energy. There are exactly n shortcuts in Mike's city, the ith of them allows walking from intersection i to intersection ai (i ≤ ai ≤ ai + 1) (but not in the opposite direction), thus there is exactly one shortcut starting at each intersection. Formally, if Mike chooses a sequence p1 = 1, p2, ..., pk then for each 1 ≤ i < k satisfying pi + 1 = api and api ≠ pi Mike will spend only 1 unit of energy instead of |pi - pi + 1| walking from the intersection pi to intersection pi + 1. For example, if Mike chooses a sequence p1 = 1, p2 = ap1, p3 = ap2, ..., pk = apk - 1, he spends exactly k - 1 units of total energy walking around them.
Before going on his adventure, Mike asks you to find the minimum amount of energy required to reach each of the intersections from his home. Formally, for each 1 ≤ i ≤ n Mike is interested in finding minimum possible total energy of some sequence p1 = 1, p2, ..., pk = i.
Input
The first line contains an integer n(1 ≤ n ≤ 200 000) — the number of Mike's city intersection.
The second line contains n integers a1, a2, ..., an(i ≤ ai ≤ n , , describing shortcuts of Mike's city, allowing to walk from intersection i to intersection ai using only 1 unit of energy. Please note that the shortcuts don't allow walking in opposite directions (from ai to i).
Output
In the only line print n integers m1, m2, ..., mn, where mi denotes the least amount of total energy required to walk from intersection 1 to intersection i.
Sample Input
Input
3
2 2 3
Output
0 1 2
Input
5
1 2 3 4 5
Output
0 1 2 3 4
Input
7
4 4 4 4 7 7 7
Output
0 1 2 1 2 3 3
Hint
In the first sample case desired sequences are:
1: 1; m1 = 0;
2: 1, 2; m2 = 1;
3: 1, 3; m3 = |3 - 1| = 2.
In the second sample case the sequence for any intersection 1 < i is always 1, i and mi = |1 - i|.
In the third sample case — consider the following intersection sequences:
1: 1; m1 = 0;
2: 1, 2; m2 = |2 - 1| = 1;
3: 1, 4, 3; m3 = 1 + |4 - 3| = 2;
4: 1, 4; m4 = 1;
5: 1, 4, 5; m5 = 1 + |4 - 5| = 2;
6: 1, 4, 6; m6 = 1 + |4 - 6| = 3;
7: 1, 4, 5, 7; m7 = 1 + |4 - 5| + 1 = 3.
题意:
图中任意两点之间距离为abs(i-j),每个点有且仅有一条捷径a[i]并且距离为1;
求从1开始遍历所有点的最小代价;
题解:
只考虑相邻两点和捷径这三条长度为1的边,最短路即可;
或者直接考虑上述三条边跑一遍bfs.
(以下代码为bfs法)
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<set>
#include<list>
#define LL long long
#define maxn 210000
#define inf 0x3f3f3f3f
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
int n;
queue<int> myq;
int dis[maxn];
int a[maxn];
int main(int argc, char const *argv[])
{
//IN;
while(scanf("%d",&n) != EOF)
{
for(int i=1; i<=n; i++) scanf("%d", &a[i]);
fill(dis, dis+maxn, inf);
while(!myq.empty()) myq.pop();
myq.push(1); dis[1] = 0;
while(!myq.empty()) {
int cur = myq.front(); myq.pop();
if(cur+1<=n && dis[cur+1] > dis[cur]+1){
dis[cur+1] = dis[cur]+1;
myq.push(cur+1);
}
if(cur-1>0 && dis[cur-1] > dis[cur]+1){
dis[cur-1] = dis[cur]+1;
myq.push(cur-1);
}
if(dis[a[cur]] > dis[cur]+1){
dis[a[cur]] = dis[cur]+1;
myq.push(a[cur]);
}
}
for(int i=1; i<=n; i++)
printf("%d ", dis[i]);;
}
return 0;
}
CodeForces 689B Mike and Shortcuts (bfs or 最短路)的更多相关文章
- CodeForces 689B Mike and Shortcuts (BFS or 最短路)
题目链接:http://codeforces.com/problemset/problem/689/B 题目大意: 留坑 明天中秋~
- Codeforces 689B. Mike and Shortcuts SPFA/搜索
B. Mike and Shortcuts time limit per test: 3 seconds memory limit per test: 256 megabytes input: sta ...
- codeforces 689B Mike and Shortcuts 最短路
题目大意:给出n个点,两点间的常规路为双向路,路长为两点之间的差的绝对值,第二行为捷径,捷径为单向路(第i个点到ai点),距离为1.问1到各个点之间的最短距离. 题目思路:SPFA求最短路 #incl ...
- codeforces 689 Mike and Shortcuts(最短路)
codeforces 689 Mike and Shortcuts(最短路) 原题 任意两点的距离是序号差,那么相邻点之间建边即可,同时加上题目提供的边 跑一遍dijkstra可得1点到每个点的最短路 ...
- codeforces 689B B. Mike and Shortcuts(bfs)
题目链接: B. Mike and Shortcuts time limit per test 3 seconds memory limit per test 256 megabytes input ...
- Codeforces Round #361 (Div. 2) B. Mike and Shortcuts bfs
B. Mike and Shortcuts 题目连接: http://www.codeforces.com/contest/689/problem/B Description Recently, Mi ...
- Codeforces Round #361 (Div. 2)——B. Mike and Shortcuts(BFS+小坑)
B. Mike and Shortcuts time limit per test 3 seconds memory limit per test 256 megabytes input standa ...
- hdu4135-Co-prime & Codeforces 547C Mike and Foam (容斥原理)
hdu4135 求[L,R]范围内与N互质的数的个数. 分别求[1,L]和[1,R]和n互质的个数,求差. 利用容斥原理求解. 二进制枚举每一种质数的组合,奇加偶减. #include <bit ...
- codeforces 547E Mike and Friends
codeforces 547E Mike and Friends 题意 题解 代码 #include<bits/stdc++.h> using namespace std; #define ...
随机推荐
- 【C#设计模式——创建型模式】工场方法模式
工场方法模式对简单工场模式进行了乔庙的扩展,不是用一个专门的类来决定实例化哪一个子类.相反,超类把这种决定延迟到每个子类.这种模式实际上没有决策点,就是没有直接选择一个子类实例化的决策. 看书上的例子 ...
- 加密解密(11)HMAC-在sha1,md5基础上加密
HMAC: Hash-based Message Authentication Code http://baike.sogou.com/v10977193.htm http://www.baike.c ...
- Windows 7下配置JDK环境变量
安装jdk1.8版本(下载链接:http://www.oracle.com/technetwork/java/javase/downloads/jdk8-downloads-2133151.html) ...
- 使用 Oracle GoldenGate 在 Microsoft SQL Server 和 Oracle Database 之间复制事务
使用 Oracle GoldenGate 在 Microsoft SQL Server 和 Oracle Database 之间复制事务 作者:Nikolay Manchev 分步构建一个跨这些平台的 ...
- 信号量及PV原语
操作系统中进程互斥和同步的实现的一个最基本的方方是使用信号量和PV原语. 信号量S的物理意义:当S≥0的时候表示,某个资源可以使用的数量,当S<0的时候,其绝对值表示等待某个资源的进程数. 一般 ...
- 面试题_48_to_65_Java 集合框架的面试题
这部分也包含数据结构.算法及数组的面试问题 48) List.Set.Map 和 Queue 之间的区别(答案)List 是一个有序集合,允许元素重复.它的某些实现可以提供基于下标值的常量访问时间,但 ...
- decode-string(挺麻烦的)
Java String作为参数传参是不会改变的,这个与常识的感觉不同. public String decodeString(String s) { s = ""; return ...
- WIN7开无线
可以的,WIN7自身就带有这个功能的,不过默认是不开启的,也可以下载一个软件来弄.1.请却仍您使用的操作系统是微软的Windows 7或者Windows server 2008 R2,正版盗版皆可.( ...
- Java [Leetcode 111]Minimum Depth of Binary Tree
题目描述: Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along th ...
- java 访问器方法中对象引用的问题
"注意不要编写返回引用可变对象的访问器方法".因为会破坏类的封装性,引用的内容可能会被改变,产生业务逻辑上的错误. 什么是可变对象? 先要搞清楚java中值传递和引用传递的问题,总结如下: 1.对象就 ...