Xenia and Weights
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia has a set of weights and pan scales. Each weight has an integer weight from 1 to 10 kilos. Xenia is going to play with scales and weights a little. For this, she puts weights on the scalepans, one by one. The first weight goes on the left scalepan, the second weight goes on the right scalepan, the third one goes on the left scalepan, the fourth one goes on the right scalepan and so on. Xenia wants to put the total of m weights on the scalepans.

Simply putting weights on the scales is not interesting, so Xenia has set some rules. First, she does not put on the scales two consecutive weights of the same weight. That is, the weight that goes i-th should be different from the(i + 1)-th weight for any i (1 ≤ i < m). Second, every time Xenia puts a weight on some scalepan, she wants this scalepan to outweigh the other one. That is, the sum of the weights on the corresponding scalepan must be strictly greater than the sum on the other pan.

You are given all types of weights available for Xenia. You can assume that the girl has an infinite number of weights of each specified type. Your task is to help Xenia lay m weights on ​​the scales or to say that it can't be done.

Input

The first line contains a string consisting of exactly ten zeroes and ones: the i-th (i ≥ 1) character in the line equals "1" if Xenia has i kilo weights, otherwise the character equals "0". The second line contains integer m (1 ≤ m ≤ 1000).

Output

In the first line print "YES", if there is a way to put m weights on the scales by all rules. Otherwise, print in the first line "NO". If you can put m weights on the scales, then print in the next line m integers — the weights' weights in the order you put them on the scales.

If there are multiple solutions, you can print any of them.

数据量小,直接搜索即可

 #include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <utility>
#include <cstdio>
#include <cstring> using namespace std; char s[];
vector<int> ans;
int m; bool dfs(int n, int rw, int lw, bool rig)
{
if(n == ) return true;
for(int i = ; i <= ; i++){
if(s[i] == ''){
if(!ans.empty() && ans.back() == i) continue;
else{
if(rig){
if(rw + i > lw){
ans.push_back(i);
if(dfs(n - , rw + i, lw, false)) return true;
ans.pop_back();
}
} else {
if(lw + i > rw){
ans.push_back(i);
if(dfs(n - , rw, lw + i, true)) return true;
ans.pop_back();
}
}
}
}
}
return false;
} int main()
{
while(scanf("%s", s + ) != EOF){
scanf("%d", &m);
ans.clear();
if(dfs(m, , , true)){
puts("YES");
vector<int>::iterator it = ans.begin();
printf("%d", *it);
for(it++; it != ans.end(); it++)
printf(" %d", *it);
puts("");
}
else puts("NO");
}
return ;
}

Xenia and Weights(深度优先搜索)的更多相关文章

  1. 深度优先搜索(DFS)

    [算法入门] 郭志伟@SYSU:raphealguo(at)qq.com 2012/05/12 1.前言 深度优先搜索(缩写DFS)有点类似广度优先搜索,也是对一个连通图进行遍历的算法.它的思想是从一 ...

  2. 初涉深度优先搜索--Java学习笔记(二)

    版权声明: 本文由Faye_Zuo发布于http://www.cnblogs.com/zuofeiyi/, 本文可以被全部的转载或者部分使用,但请注明出处. 上周学习了数组和链表,有点基础了解以后,这 ...

  3. 挑战程序2.1.4 穷竭搜索>>深度优先搜索

      深度优先搜索DFS,从最开始状态出发,遍历一种状态到底,再回溯搜索第二种. 题目:POJ2386  思路:(⊙v⊙)嗯  和例题同理啊,从@开始,搜索到所有可以走到的地方,把那里改为一个值(@或者 ...

  4. 回溯 DFS 深度优先搜索[待更新]

      首先申明,本文根据微博博友 @JC向北 微博日志 整理得到,本文在这转载已经受作者授权!   1.概念   回溯算法 就是 如果这个节点不满足条件 (比如说已经被访问过了),就回到上一个节点尝试别 ...

  5. 总结A*,Dijkstra,广度优先搜索,深度优先搜索的复杂度比较

    广度优先搜索(BFS) 1.将头结点放入队列Q中 2.while Q!=空 u出队 遍历u的邻接表中的每个节点v 将v插入队列中 当使用无向图的邻接表时,复杂度为O(V^2) 当使用有向图的邻接表时, ...

  6. [codeforces 339]C. Xenia and Weights

    [codeforces 339]C. Xenia and Weights 试题描述 Xenia has a set of weights and pan scales. Each weight has ...

  7. 深度优先搜索(DFS)

    定义: (维基百科:https://en.wikipedia.org/wiki/Depth-first_search) 深度优先搜索算法(Depth-First-Search),是搜索算法的一种.是沿 ...

  8. 图的遍历之深度优先搜索(DFS)

    深度优先搜索(depth-first search)是对先序遍历(preorder traversal)的推广.”深度优先搜索“,顾名思义就是尽可能深的搜索一个图.想象你是身处一个迷宫的入口,迷宫中的 ...

  9. HDU 1241 Oil Deposits DFS(深度优先搜索) 和 BFS(广度优先搜索)

    Oil Deposits Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total ...

随机推荐

  1. PAIP.MYSQL SLEEP 连接太多解决

    PAIP.MYSQL SLEEP 连接太多解决 作者Attilax  艾龙,  EMAIL:1466519819@qq.com  来源:attilax的专栏 地址:http://blog.csdn.n ...

  2. paip.java 线程无限wait的解决

    paip.java  线程无限wait的解决 jprofl>threads>thread dump> 查看棉线程执行的code stack... 估计是.比如.BlockingQue ...

  3. paip.log4j兼容linux windows 路径设置

    paip.log4j兼容linux windows 路径设置 作者Attilax  艾龙,  EMAIL:1466519819@qq.com  来源:attilax的专栏 地址:http://blog ...

  4. ServiceStack Web Service 创建与调用简单示列

    目录 ServiceStack 概念 ServiceStack Web Service 创建与调用简单示列 上篇文章介绍了ServiceStack是什么,本章进入主题,如何快速简单的搭建Service ...

  5. O2O已死?不!美团点评们迎来新风口

    当年的千团大战,巅峰时期曾涌入了5000多家团购网站,刘旷本人也参与了此次团购大战.而就在当时很多人都唱衰团购的时候,美团和大众点评却最终脱颖而出,市值一路飙升,人人网旗下的糯米网因为卖给了百度,也得 ...

  6. asp.net对cookie的操作

    创建cookie: HttpCookie cookie = new HttpCookie("CurrentUser"); //创建一个名称为CurrentUser 的cookie对 ...

  7. Ubuntu14.04.1 阿里apt源

    deb http://mirrors.aliyun.com/ubuntu/ trusty main restricted universe multiversedeb http://mirrors.a ...

  8. Android中的IOC框架,完全注解方式就可以进行UI绑定和事件绑定

    转载请注明出处:http://blog.csdn.net/blog_wang/article/details/38468547 相信很多使用过Afinal和Xutils的朋友会发现框架中自带View控 ...

  9. spring发送邮件(多人接收或抄送多少带附件发送)

    系统中的附件分享功能界面 抄送多个效果图 多个接收者效果图 抄送多人带附件源码 多个接收者带附件源码

  10. WPF窗体的命令绑定

    方法一:使用代码 <WpfUI:View.CommandBindings> <CommandBinding Command="Help" CanExecute=& ...