hdu 5233 Gunner II （map的简单用法）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5233

Gunner II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1752    Accepted Submission(s): 639

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

 题目大意：上面一排数字，下面一排数字，找出下面一排数字在上面一排中的位置，如果有多个，输出优先靠左边的，然后删掉其数字。如果不存在输出-1

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <vector>

using namespace std;

int h[100005], q[100005];

map<int,vector<int> > mp;

int main ()
{
    int m, n;
    while(~scanf("%d %d", &n, &m))
    {
        mp.clear();
        for(int i=1; i<=n; i++)
        {
            scanf("%d", &h[i]);
        }
        for(int i=n; i>=1; i--)
        {
            mp[h[i]].push_back(i);
        }
        for(int i=1; i<=m; i++)
        {
            scanf("%d", &q[i]);
            if(mp[q[i]].size()>=1)
            {
                printf("%d
",mp[q[i]].back());
                mp[q[i]].pop_back();
            }
            else
                printf("-1
");

        }
    }

    return 0;
}