Gunner II

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1433    Accepted Submission(s): 540 **

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

Hint

Huge input, fast IO is recommended.

Source

BestCoder Round #42

因为一共才1E5的数据，而高度有1E9

所以考虑离散化

关于离散化的内容，这篇博客讲得很好。

http://blog.csdn.net/axuan_k/article/details/45954561

我是使用了第三种map+set的方法。。。

离散化大概是因为数据太大下表存不下。。。

然后map的key 值没有范围？所以实现了离散化（是这样嘛？？？）

/*************************************************************************
	> File Name: code/bc/#42/B.cpp
	> Author: 111qqz
	> Email: rkz2013@126.com 
	> Created Time: 2015年08月01日 星期六 02时47分54秒
 ************************************************************************/

 1#include<iostream>
 2#include<iomanip>
 3#include<cstdio>
 4#include<algorithm>
 5#include<cmath>
 6#include<cstring>
 7#include<string>
 8#include<map>
 9#include<set>
10#include<queue>
11#include<vector>
12#include<stack>
13#define y0 abc111qqz
14#define y1 hust111qqz
15#define yn hez111qqz
16#define j1 cute111qqz
17#define tm crazy111qqz
18#define lr dying111qqz
19using namespace std;
20#define REP(i, n) for (int i=0;i<int(n);++i)  
21typedef long long LL;
22typedef unsigned long long ULL;
23const int inf = 0x7fffffff;
24const int N=2E5+7;
25int n,m;
26map<int,int>mp;
27set<int>se[N];
28int main()
29{
30    while (scanf("%d %d",&n,&m)!=EOF)
31    {
32	for ( int i = 1 ; i <= n ; i++ )
33	    se[i].clear();
34	mp.clear();
35	int  t = 0;
36	int x;
37	for ( int i = 1 ; i <= n ; i++ )
38	{
39	    scanf("%d",&x);
40	    if (!mp[x]) mp[x]=++t;
41	    se[mp[x]].insert(i);
42	}
43	for ( int i = 1 ; i <= m ; i++)
44	{
45	    scanf("%d",&x);
46	    if (se[mp[x]].size()==0)
47	    {
48		puts("-1");
49	    }
50	    else
51	    {
52		printf("%d\n",*se[mp[x]].begin());
53		se[mp[x]].erase(se[mp[x]].begin());
54	    }
55	}

1    }
2	return 0;
3}

还有一种写法，貌似要快一点，但是空间用的多一些：

/*************************************************************************
	> File Name: code/bc/#42/BB.cpp
	> Author: 111qqz
	> Email: rkz2013@126.com 
	> Created Time: 2015年08月01日 星期六 09时30分45秒
 ************************************************************************/

 1#include<iostream>
 2#include<iomanip>
 3#include<cstdio>
 4#include<algorithm>
 5#include<cmath>
 6#include<cstring>
 7#include<string>
 8#include<map>
 9#include<set>
10#include<queue>
11#include<vector>
12#include<stack>
13#define y0 abc111qqz
14#define y1 hust111qqz
15#define yn hez111qqz
16#define j1 cute111qqz
17#define tm crazy111qqz
18#define lr dying111qqz
19using namespace std;
20#define REP(i, n) for (int i=0;i<int(n);++i)  
21typedef long long LL;
22typedef unsigned long long ULL;
23const int inf = 0x7fffffff;
24map<int,set<int> >mp;
25int m,n;
26int main()
27{
28    while (scanf("%d %d",&n,&m)!=EOF)
29    {
30	mp.clear();
31	for ( int i = 1;  i <= n ; i++ )
32	{
33	    int tmpx;
34	    scanf("%d",&tmpx);
35	    mp[tmpx].insert(i);
36	}
37	for ( int i = 1 ; i <= m ; i++ )
38	{
39	    int q;
40	    scanf("%d",&q);
41	    if (mp[q].size()==0)
42	    {
43		puts("-1");
44	    }
45	    else
46	    {
47		printf("%d\n",*mp[q].begin());
48		mp[q].erase(mp[q].begin());
49	    }
50	}
51    }

	return 0;
}