hdu5233 Gunner II

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

这道题因为删除方法不对，一直T,终于改用vis后，900多ms水过去了。。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
struct node{
	int id,num;
}a[100006];
int n,vis[100006];

bool cmp(node a,node b){
	if(a.num==b.num)return a.id>b.id;
	return a.num<b.num;
}
int q[100006];

int find(int x,int l,int r)
{
	int mid,j;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(a[mid].num==x)break;
		if(a[mid].num>x)r=mid-1;
		else l=mid+1;
	}
	j=mid;
	if(vis[j]==1){
		j--;
		while(1){
			if(vis[j]==0)break;
			j--;
		}
		return j;
	}
	else{
		j++;
		while(1){
			if(j>n || vis[j]==1 || a[j].num>x){
				j--;break;
			}
			j++;
		}
		return j;
	}
}

int main()
{
	int m,i,j,k;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		map<int,int>hash;
		hash.clear();
		for(i=1;i<=n;i++){
			vis[i]=0;
			scanf("%d",&a[i].num);
			a[i].id=i;
			hash[a[i].num]++;
		}
		sort(a+1,a+1+n,cmp);
		for(i=1;i<=m;i++){
			scanf("%d",&q[i]);
			if(hash[q[i]]==0){
				printf("-1\n");continue;
			}
			else{
				hash[q[i]]--;
				j=find(q[i],1,n);
				vis[j]=1;
				printf("%d\n",a[j].id);
			}
		}
	}
	return 0;
}