HDU5233
Gunner II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1244 Accepted Submission(s): 486
Problem Description
Long long ago, there was a gunner whose name is Jack. He likes to go hunting very much. One day he go to the grove. There are n birds and n trees. The i-th bird stands on the top of the i-th tree. The trees stand in straight line from left to the right. Every
tree has its height. Jack stands on the left side of the left most tree. When Jack shots a bullet in height H to the right, the nearest bird which stands in the tree with height H will falls.
Jack will shot many times, he wants to know which bird will fall during each shot.
Input
There are multiple test cases (about 5), every case gives n, m in the first line, n indicates there are n trees and n birds, m means Jack will shot m times.
In the second line, there are n numbers h[1],h[2],h[3],…,h[n] which describes the height of the trees.
In the third line, there are m numbers q[1],q[2],q[3],…,q[m] which describes the height of the Jack’s shots.
Please process to the end of file.
[Technical Specification]
All input items are integers.
1<=n,m<=100000(10^5)
1<=h[i],q[i]<=1000000000(10^9)
Output
For each q[i], output an integer in a single line indicates the id of bird Jack shots down. If Jack can’t shot any bird, just output -1.
The id starts from 1.
Sample Input
5 5
1 2 3 4 1
1 3 1 4 2
Sample Output
1
3
5
4
2
//这题主要思路就是利用一个结构体将高度与序号记录下来
//再利用一个数组将高度排序而且标记高度出现的顺序 #include <stdio.h>
#include <algorithm>
using namespace std;
int flag[100010],h[100010]; struct bird
{
int n,num;
}p[100010]; bool cmp(const bird &a,const bird &b)
{
return a.n!=b.n? a.n<b.n:a.num<b.num;
} int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
for(int i=0;i<n;i++ )
{
scanf("%d",&p[i].n);
p[i].num=i+1;
}
sort(p,p+n,cmp);
for(int i=0;i<n;i++)
{
flag[i]=i; //此时的flag数组是标记高度第一次出现的位置
h[i]=p[i].n;
}
while(m--)
{
int x;
scanf("%d",&x);
int q=lower_bound(h,h+n,x)-h; //查询到的也是第一次出现的次数
if(h[flag[q]]!=x)
{
printf("-1\n");
continue;
}
else
printf("%d\n",p[flag[q]].num);
flag[q]++; //因为查询过一次了 所以递增到下一个高度
}
}
return 0;
}
HDU5233的更多相关文章
- hdu5233 Gunner II
Problem Description Long long ago, there was a gunner whose name is Jack. He likes to go hunting ver ...
- Beatcoder#39+#41+#42
HDU5211 思路: 倒着更新每个数的约数,更新完要把自己加上,以及1的情况? //#include <bits/stdc++.h> #include<iostream> # ...
随机推荐
- CCDirector导演类
CCDirector类是Cocos2D-x游戏引擎的核心.它用来创建而且控制着屏幕的显示,同一时候控制场景的显示时间和显示方式. 在整个游戏里一般仅仅有一个导演.游戏的開始.结束.暂停都会调用CCDi ...
- HTML5开发移动web应用——Sencha Touch篇(8)
DataView是Sencha Touch中最重要的组件,用于数据的可视化.数据可视化的重要性不言而喻,能够讲不论什么数据以形象的方式展示给用户. 眼下,怎样更好地可视化是很多公司或框架都在追求的. ...
- 英语发音规则---T字母
英语发音规则---T字母 一.总结 一句话总结: 1.T一般发[t]? ten [ten] num.十 letter [ˈletə(r)] n.信; 证书 meet [mi:t] vt.& v ...
- 高斯混合模型Gaussian Mixture Model (GMM)——通过增加 Model 的个数,我们可以任意地逼近任何连续的概率密分布
从几何上讲,单高斯分布模型在二维空间应该近似于椭圆,在三维空间上近似于椭球.遗憾的是在很多分类问题中,属于同一类别的样本点并不满足“椭圆”分布的特性.这就引入了高斯混合模型.——可以认为是基本假设! ...
- [jzoj 5661] 药香沁鼻 解题报告 (DP+dfs序)
interlinkage: https://jzoj.net/senior/#contest/show/2703/0 description: solution: 注意到这本质就是一个背包,只是选了一 ...
- MVC的一些常用特性,持续更新中。。。
1. @MvcHtmlString.Create("<option value='1'>火星</option>") //渲染Html
- Habernate配置一对一,一对多,多对多(二)
一.开篇 紧接着上篇的博客来写:http://www.cnblogs.com/WJ--NET/p/7845000.html(habernate环境的搭建) 二.配置一对一 2.1.新建客户类和公司类( ...
- Redis的配置文件详解
daemonize:如需要在后台运行,把该项的值改为yes pdifile:把pid文件放在/var/run/redis.pid,可以配置到其他地址 bind:指定redis只接收来自该IP的请求,如 ...
- 博客移至 GitHub
新博客地址: github.com/FatliTalk/blog
- LeetCode(15)3Sum
题目如下: Python代码: def threeSum(self, nums): res = [] nums.sort() for i in xrange(len(nums)-2): if i &g ...