hdu 5233 Gunner II (bc #42 B) （离散化）

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
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=2E5+7;
int n,m;
map<int,int>mp;
set<int>se[N];
int main()
{
    while (scanf("%d %d",&n,&m)!=EOF)
    {
    for ( int i = 1 ; i <= n ; i++ )
        se[i].clear();
    mp.clear();
    int  t = 0;
    int x;
    for ( int i = 1 ; i <= n ; i++ )
    {
        scanf("%d",&x);
        if (!mp[x]) mp[x]=++t;
        se[mp[x]].insert(i);
    }
    for ( int i = 1 ; i <= m ; i++)
    {
        scanf("%d",&x);
        if (se[mp[x]].size()==0)
        {
        puts("-1");
        }
        else
        {
        printf("%d
",*se[mp[x]].begin());
        se[mp[x]].erase(se[mp[x]].begin());
        }
    }

    }
    return 0;
}

 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
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
map<int,set<int> >mp;
int m,n;
int main()
{
    while (scanf("%d %d",&n,&m)!=EOF)
    {
    mp.clear();
    for ( int i = 1;  i <= n ; i++ )
    {
        int tmpx;
        scanf("%d",&tmpx);
        mp[tmpx].insert(i);
    }
    for ( int i = 1 ; i <= m ; i++ )
    {
        int q;
        scanf("%d",&q);
        if (mp[q].size()==0)
        {
        puts("-1");
        }
        else
        {
        printf("%d
",*mp[q].begin());
        mp[q].erase(mp[q].begin());
        }
    }
    }
  
    return 0;
}