hdu 5233 Gunner II【STL应用】【水题】

Gunner II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1607    Accepted Submission(s): 604

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 51 2 3 4 11 3 1 4 2

 

Sample Output

13542
Hint
Huge input, fast IO is recommended.

二位数组是永远开不了那么大的，看到数据就应该知道这一点，所以要应用map，但是这里是一个一个干掉这个高度的鸟的，所以这里应用map+queue的组合就能AC了~
但是不知道为什么交G++是MLE，但是交C++就能AC~

#include<stdio.h>#include<string.h>#include<map>#include<queue>using namespace std;int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        map<int,queue<int> >map;        for(int i=1;i<=n;i++)        {            int k;            scanf("%d",&k);            map[k].push(i);        }        for(int i=1;i<=m;i++)        {            int k;            scanf("%d",&k);            if(map[k].size()==0)            {                printf("-1\n");            }            else            {                printf("%d\n",map[k].front());                map[k].pop();            }        }    }}