hdoj--5233--Gunner II(map+queue&&二分)



Gunner II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1724    Accepted Submission(s): 631


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
 感觉二分写的应该对的啊,但是一直wa
错误代码:

#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <stack>
#include<iostream>
using namespace std;
struct Node
{
	int num,hight;
	bool flg;
}a[110000];
bool cmp(Node x,Node y)
{
	if(x.hight == y.hight )return x.num<y.num;
	return x.hight<y.hight;
}
int b_search(int val,int len)
{
	int l=0,r=len-1,mid;
	while(l<=r)
	{
		mid=(l+r)/2;
		if(a[mid].hight<val)l=mid+1;
		else if(a[mid].hight == val && a[mid].flg == false)l=mid+1;
		else r=mid-1;
	}
	return l;
}
int main()
{
	int m,n,i,j,cy,ans,ji;
	while(~scanf("%d %d",&n,&m))
	{
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i].hight);
			a[i].num=i+1;
			a[i].flg=true;
		}
		sort(a,a+n,cmp);
		for(i=0;i<m;i++)
		{
			scanf("%d",&ji);
			ans=b_search(ji,n);
			a[ans].flg=false;
			if(a[ans].hight==j)
			printf("-1
");
			else
			{
				a[ans].flg=false;
				cout<<a[ans].num<<endl;
			}
		}
	}
	return 0;
}
 
map+queue
#include<cstdio>
#include<map>
#include<list>
#include<queue>
#include<string>
#include<algorithm>
using namespace std; 
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m))
	{
		map<int,queue<int> >mp;
		int tmd;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&tmd);
			mp[tmd].push(i);
		}
		while(m--)
		{
			scanf("%d",&tmd);
			queue<int> &p=mp[tmd];
			if(p.empty()) printf("-1
");
			else 
			{
				printf("%d
",p.front());
				p.pop();
			}
		}
	}
	return 0;
}

 
原文地址:https://www.cnblogs.com/playboy307/p/5273430.html