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
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标存不下……
然后map的key 值没有范围？所以实现了离散化（是这样嘛？？？）
代码实现
1/*************************************************************************
2 > File Name: code/bc/#42/B.cpp
3 > Author: 111qqz
4 > Email: rkz2013@126.com
5 > Created Time: 2015年08月01日 星期六 02时47分54秒
6 ************************************************************************/
7
8#include<iostream>
9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#define y0 abc111qqz
21#define y1 hust111qqz
22#define yn hez111qqz
23#define j1 cute111qqz
24#define tm crazy111qqz
25#define lr dying111qqz
26using namespace std;
27#define REP(i, n) for (int i=0;i<int(n);++i)
28typedef long long LL;
29typedef unsigned long long ULL;
30const int inf = 0x7fffffff;
31const int N=2E5+7;
32int n,m;
33map<int,int>mp;
34set<int>se[N];
35int main()
36{
37 while (scanf("%d %d",&n,&m)!=EOF)
38 {
39 for ( int i = 1 ; i <= n ; i++ )
40 se[i].clear();
41 mp.clear();
42 int t = 0;
43 int x;
44 for ( int i = 1 ; i <= n ; i++ )
45 {
46 scanf("%d",&x);
47 if (!mp[x]) mp[x]=++t;
48 se[mp[x]].insert(i);
49 }
50 for ( int i = 1 ; i <= m ; i++)
51 {
52 scanf("%d",&x);
53 if (se[mp[x]].size()==0)
54 {
55 puts("-1");
56 }
57 else
58 {
59 printf("%d\n",*se[mp[x]].begin());
60 se[mp[x]].erase(se[mp[x]].begin());
61 }
62 }
63
64 }
65 return 0;
66}还有一种写法，貌似要快一点，但是空间用的多一些：
代码实现
1/*************************************************************************
2 > File Name: code/bc/#42/BB.cpp
3 > Author: 111qqz
4 > Email: rkz2013@126.com
5 > Created Time: 2015年08月01日 星期六 09时30分45秒
6 ************************************************************************/
7
8#include<iostream>
9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#define y0 abc111qqz
21#define y1 hust111qqz
22#define yn hez111qqz
23#define j1 cute111qqz
24#define tm crazy111qqz
25#define lr dying111qqz
26using namespace std;
27#define REP(i, n) for (int i=0;i<int(n);++i)
28typedef long long LL;
29typedef unsigned long long ULL;
30const int inf = 0x7fffffff;
31map<int,set<int> >mp;
32int m,n;
33int main()
34{
35 while (scanf("%d %d",&n,&m)!=EOF)
36 {
37 mp.clear();
38 for ( int i = 1; i <= n ; i++ )
39 {
40 int tmpx;
41 scanf("%d",&tmpx);
42 mp[tmpx].insert(i);
43 }
44 for ( int i = 1 ; i <= m ; i++ )
45 {
46 int q;
47 scanf("%d",&q);
48 if (mp[q].size()==0)
49 {
50 puts("-1");
51 }
52 else
53 {
54 printf("%d\n",*mp[q].begin());
55 mp[q].erase(mp[q].begin());
56 }
57 }
58 }
59
60 return 0;
61}