hdu 5233 Gunner II（数据离散化存储）

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
   
   
   
   

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

用map<int,queue>会超时。还是老老实实地用数组吧。。。或者可以用map<int,int>离散化再对应queue<int>

#include <stdio.h>  
#include <string.h>  
#include <iostream>  
#include <algorithm>  
#include <vector>  
#include <queue>  
#include <stack>  
#include <set>  
#include <map>  
#include <string>  
#include <math.h>  
#include <stdlib.h>  
#include <time.h>  
using namespace std;  
const int N=1E5+5;  
struct mapp{  
    int h,p;  
}a[N];  
int n,m,p,x,h[N],f[N];  
  
  int comp(const mapp&a,const mapp&b)  
{  
    if (a.h==b.h) return a.p<b.p;  
    return a.h<b.h;  
}  
  
int main()  
{  
    while (cin>>n>>m)  
    {  
        for (int i=1;i<=n;i++) scanf("%d",&a[i].h),a[i].p=i;  
        sort(a+1,a+n+1,comp);  
        for (int i=1;i<=n;i++) f[i]=i,h[i]=a[i].h; //h[]以从小到大存高度，f[]能用到的部分表示每种高度的第一位置 
        for (int i=1;i<=m;i++)  
        {  
            scanf("%d",&x);  
            p=lower_bound(h+1,h+n+1,x)-h;  
            if (h[f[p]]!=x) 
			{ 
				printf("-1\n"); 
				continue; 
			}  
            printf("%d\n",a[f[p]].p);  
            f[p]++; //由该高度的第一位置转到第二位置 
        }  
    }  
    return 0; 
}  

或：

#include <stdio.h>
#include <string.h>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
typedef __int64 ll;

const int maxn=100000+10;
queue<int> que[maxn];

int main()
{
    int i,n,j,m,t;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(i=1;i<=n;i++)
            while(!que[i].empty())
                que[i].pop();
        map<int,int> mp;
        int cnt=1;
        for(i=1;i<=n;i++){
            scanf("%d",&t);
            int tt=mp[t];
            if(tt)
                que[tt].push(i);
            else {
                    mp[t]=cnt++;
                que[cnt-1].push(i);
            }
        }
        while(m--){
            scanf("%d",&t);
            int tt=mp[t];
            if(que[tt].empty())
                printf("-1\n");
            else {
                printf("%d\n",que[tt].front());
                que[tt].pop();
            }
        }
    }
    return 0;
}