hdu 4995(枚举)

Revenge of kNN

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

[align=left]Problem Description[/align]
In pattern recognition, the k-Nearest Neighbors algorithm (or k-NN for short) is a non-parametric method used for classification and regression. In both cases, the input consists of the k closest training examples in the feature space.

In k-NN regression, the output is the property value for the object. This value is the average of the values of its k nearest neighbors.

---Wikipedia

Today, kNN takes revenge on you. You have to handle a kNN case in one-dimensional coordinate system. There are N points with a position Xi and value Vi. Then there are M kNN queries for point with index i, recalculate its value by averaging the values its k-Nearest
Neighbors. Note you have to replace the value of i-th point with the new calculated value. And if there is a tie while choosing k-Nearest Neighbor, choose the one with the minimal index first.
 

[align=left]Input[/align]
The first line contains a single integer T, indicating the number of test cases.

Each test case begins with three integers N, M and K, in which K indicating the number of k-Nearest Neighbors. Then N lines follows, each line contains two integers Xi and Vi. Then M lines with the queried index Qi follows.

[Technical Specification]

1. 1 <= T <= 5

2. 2<=N<= 100 000, 1<=M<=100 000

3. 1 <= K <= min(N – 1, 10)

4. 1 <= Vi <= 1 000

5. 1 <= Xi <= 1 000 000 000, and no two Xi are identical.

6. 1 <= Qi <= N
 

[align=left]Output[/align]
For each test case, output sum of all queries rounded to six fractional digits.
 

[align=left]Sample Input[/align]

1
5 3 2
1 2
2 3
3 6
4 8
5 8
2
3
4

 

[align=left]Sample Output[/align]

17.000000
HintFor the first query, the 2-NN for point 2 is point 1 and 3, so the new value is (2 + 6) / 2 = 4.
For the second query, the 2-NN for point 3 is point 2 and 4, and the value of point 2 is changed to 4 by the last query, so the new value is (4 + 8) / 2 = 6.

Huge input, faster I/O method is recommended.解题思路：读懂题意，暴力枚举即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int maxn = 100005;
struct Point
{
__int64 x;
int id;
double value;
}p[maxn];
int n,m,k,cnt,idx[maxn];

bool cmp(Point a,Point b)
{
return a.x < b.x;
}

int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&k);
for(int i = 1; i <= n; i++)
{
scanf("%I64d%lf",&p[i].x,&p[i].value);
p[i].id = i;
}
sort(p+1,p+1+n,cmp);
for(int i = 1; i <= n; i++)
idx[p[i].id] = i;
double sum = 0;
while(m--)
{
int q;
double ans = 0;
scanf("%d",&q);
int pos = idx[q];
int l = pos - 1, r = pos + 1;
cnt = 0;
while(cnt < k)
{
if(l >= 1 && r <= n)
{
if(p[pos].x - p[l].x < p[r].x - p[pos].x)
{
ans += p[l].value;
l--;
}
else if(p[pos].x - p[l].x > p[r].x - p[pos].x)
{
ans += p[r].value;
r++;
}
else if(p[l].id < p[r].id)
{
ans += p[l].value;
l--;
}
else if(p[l].id > p[r].id)
{
ans += p[r].value;
r++;
}
}
else if(l == 0)
{
ans += p[r].value;
r++;
}
else if(r == n + 1)
{
ans += p[l].value;
l--;
}
cnt++;
}
p[pos].value = ans / k;
sum += ans / k;
}
printf("%.6f\n",sum);
}
return 0;
}