POJ 2182&& POJ 2828：Lost Cows 从后往前 线段树

Lost Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10544		Accepted: 6754

Description

N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'watering hole' and drank a few too many beers before dinner. When it was time to line up for their evening meal, they did
not line up in the required ascending numerical order of their brands.

Regrettably, FJ does not have a way to sort them. Furthermore, he's not very good at observing problems. Instead of writing down each cow's brand, he determined a rather silly statistic: For each cow in line, he knows the number of cows that precede that cow
in line that do, in fact, have smaller brands than that cow.

Given this data, tell FJ the exact ordering of the cows.

Input

* Line 1: A single integer, N

* Lines 2..N: These N-1 lines describe the number of cows that precede a given cow in line and have brands smaller than that cow. Of course, no cows precede the first cow in line, so she is not listed. Line 2 of the input describes the number of preceding cows
whose brands are smaller than the cow in slot #2; line 3 describes the number of preceding cows whose brands are smaller than the cow in slot #3; and so on.

Output

* Lines 1..N: Each of the N lines of output tells the brand of a cow in line. Line #1 of the output tells the brand of the first cow in line; line 2 tells the brand of the second cow; and so on.
Sample Input

5
1
2
1
0

Sample Output

2
4
5
3
1

有N头牛站成一列，每头牛只能看见自己前面比自己brand小的牛，这个数量给你了。问你每头牛的brand分别是多少。

就是查空位数量，如果有空位被占了，那么就往后面延。。。

只不过用了线段树的形式，以线段树的形式来找空位。len代表空位的数量。

代码：

#include <iostream>  
#include <algorithm>  
#include <cmath>  
#include <vector>  
#include <string>  
#include <cstring>  
#pragma warning(disable:4996)  
using namespace std;

struct no
{
	int L,R;
	int len;
}tree[32005];

void buildtree(int root,int L,int R)
{
	tree[root].L=L;
	tree[root].R=R;

	tree[root].len = R-L+1;

	if(L!=R)
	{
		int mid = (L+R)>>1;
		buildtree((root<<1)+1,L,mid);
		buildtree((root<<1)+2,mid+1,R);
	}
}

int query(int root,int num)
{
	tree[root].len--;
	if(tree[root].L==tree[root].R)
	{
		return tree[root].L;
	}
	if(tree[(root<<1)+1].len >= num)
	{
		return query((root<<1)+1,num);
	}
	else
	{
		return query((root<<1)+2,num-tree[(root<<1)+1].len);
	}
}

int n;
int val[8005];
int ans[8005];

int main()
{
	int i;
	while(scanf("%d",&n)!=EOF)
	{
		buildtree(0,1,n);
		val[1]=0;

		for(i=2;i<=n;i++)
		{
			scanf("%d",val+i);
		}
		for(i=n;i>=1;i--)
		{
			ans[i]=query(0,val[i]+1);
		}
		for(i=1;i<=n;i++)
		{
			printf("%d\n",ans[i]);
		}
	}
	return 0;
}

POJ2828是它的姐妹题，买一送一。

代码：

#include <iostream>    
#include <algorithm>    
#include <cmath>    
#include <vector>    
#include <string>    
#include <cstring>    
#pragma warning(disable:4996)    
using namespace std;  
  
struct no  
{  
    int L,R;  
    int len;  
}tree[800005];  

int n;  
int val[200005];
int pos[200005]; 
int cnt[200005];

void buildtree(int root,int L,int R)  
{  
    tree[root].L=L;  
    tree[root].R=R;  
  
    tree[root].len = R-L+1;  
  
    if(L!=R)  
    {  
        int mid = (L+R)>>1;  
        buildtree((root<<1)+1,L,mid);  
        buildtree((root<<1)+2,mid+1,R);  
    }  
}  
  
void query(int root,int num,int val)  
{  
    tree[root].len--;  
    if(tree[root].L==tree[root].R)  
    {  
        cnt[tree[root].L]=val;
		return;
    }  
    if(tree[(root<<1)+1].len >= num)  
    {  
        query((root<<1)+1,num,val);  
    }  
    else  
    {  
        query((root<<1)+2,num-tree[(root<<1)+1].len,val);  
    }  
}  
  
int main()  
{  	
    int i;  
    while(scanf("%d",&n)!=EOF)  
    {  
        buildtree(0,1,n);  
  
        for(i=1;i<=n;i++)  
        {  
            scanf("%d%d",pos+i,val+i);  
        }  
        for(i=n;i>=1;i--)  
        {  
            query(0,pos[i]+1,val[i]);  
        }  
        for(i=1;i<=n;i++)  
        {  
			if(i==1)
                 printf("%d",cnt[i]);
			else
				 printf(" %d",cnt[i]);
        }
		printf("\n");
    }  
    return 0;  
}