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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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
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)

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

5 5
1 2 3 4 1
1 3 1 4 2

[align=left]Sample Output[/align]

1
3
5
4
2

Hint

Huge input, fast IO is recommended.

[align=left]Source[/align]
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\n",mp[q[i]].back());
mp[q[i]].pop_back();
}
else
printf("-1\n");

}
}

return 0;
}