给定2D坐标中的n个点(x i ，y i )的集合。每个点都有一定的权重w i 。任务是检查是否可以绘制45度线，以使两侧的点的权重之和相等。
例子：
Input : x1 = -1, y1 = 1, w1 = 3
x2 = -2, y2 = 1, w2 = 1
x3 = 1, y3 = -1, w3 = 4
Output : Yes
Input : x1 = 1, y1 = 1, w1 = 2
x2 = -1, y2 = 1, w2 = 1
x3 = 1, y3 = -1, w3 = 2
Output : No首先，让我们尝试解决垂直线的上述问题，即如果x = i线可以将平面分为两部分，以使每侧的权重之和相等。
可以看到，具有相同x坐标的多个点可以被视为一个权重等于具有相同x坐标的所有点的权重之和的点。
现在，遍历所有从最小x坐标到最大x坐标的x坐标。因此，创建一个数组prefix_sum [] ，它将存储权重之和，直到点x = i为止。
因此，答案可以为“是”的选项有两个：
- prefix_sum [1，2，…，i-1] = prefix_sum [i + 1，…，n]
- 或存在一个点i，使得一条线穿过之间的某处
x = i和x = i + 1以及prefix_sum [1，…，i] = prefix_sum [i + 1，…，n]，
其中prefix_sum [i，…，j]是从i到j的点的权重之和。
int is_possible = false;
for (int i = 1; i < prefix_sum.size(); i++)
if (prefix_sum[i] == total_sum - prefix_sum[i])
is_possible = true
if (prefix_sum[i-1] == total_sum - prefix_sum[i])
is_possible = true现在，要求解45度的直线，我们将每个点旋转45度。
参考：对象的2D变换或旋转
因此，在旋转45度后，指向(x，y)的点将变为((x – y)/ sqrt(2)，(x + y)/ sqrt(2))。
我们可以忽略sqrt(2)，因为它是缩放因子。同样，我们不需要关心旋转后的y坐标，因为垂直线无法区分具有相同x坐标的点。 (x，y 1 )和(x，y 2 )将位于x = k形式的右侧，左侧或任何行上。
C++
#include
using namespace std;
// Checking if a plane can be divide by a line
// at 45 degrees such that weight sum is equal
void is_partition_possible(int n, int x[],
int y[], int w[])
{
map weight_at_x;
int max_x = -2e3, min_x = 2e3;
// Rotating each point by 45 degrees and
// calculating prefix sum.
// Also, finding maximum and minimum x
// coordinates
for (int i = 0; i < n; i++) {
int new_x = x[i] - y[i];
max_x = max(max_x, new_x);
min_x = min(min_x, new_x);
// storing weight sum upto x - y point
weight_at_x[new_x] += w[i];
}
vector sum_till;
sum_till.push_back(0);
// Finding prefix sum
for (int x = min_x; x <= max_x; x++) {
sum_till.push_back(sum_till.back() +
weight_at_x[x]);
}
int total_sum = sum_till.back();
int partition_possible = false;
for (int i = 1; i < sum_till.size(); i++) {
if (sum_till[i] == total_sum - sum_till[i])
partition_possible = true;
// Line passes through i, so it neither
// falls left nor right.
if (sum_till[i - 1] == total_sum - sum_till[i])
partition_possible = true;
}
printf(partition_possible ? "YES\n" : "NO\n");
}
// Driven Program
int main()
{
int n = 3;
int x[] = { -1, -2, 1 };
int y[] = { 1, 1, -1 };
int w[] = { 3, 1, 4 };
is_partition_possible(n, x, y, w);
return 0;
} Java
import java.util.*;
// Checking if a plane can be divide by a line
// at 45 degrees such that weight sum is equal
class GFG
{
static void is_partition_possible(int n, int x[],
int y[], int w[])
{
Map weight_at_x = new HashMap();
int max_x = (int) -2e3, min_x = (int) 2e3;
// Rotating each point by 45 degrees and
// calculating prefix sum.
// Also, finding maximum and minimum x
// coordinates
for (int i = 0; i < n; i++)
{
int new_x = x[i] - y[i];
max_x = Math.max(max_x, new_x);
min_x = Math.min(min_x, new_x);
// storing weight sum upto x - y point
if(weight_at_x.containsKey(new_x))
{
weight_at_x.put(new_x, weight_at_x.get(new_x) + w[i]);
}
else
{
weight_at_x.put(new_x,w[i]);
}
//weight_at_x[new_x] += w[i];
}
Vector sum_till = new Vector<>();
sum_till.add(0);
// Finding prefix sum
for (int s = min_x; s <= max_x; s++)
{
if(weight_at_x.get(s) == null)
sum_till.add(sum_till.lastElement());
else
sum_till.add(sum_till.lastElement() +
weight_at_x.get(s));
}
int total_sum = sum_till.lastElement();
int partition_possible = 0;
for (int i = 1; i < sum_till.size(); i++)
{
if (sum_till.get(i) == total_sum - sum_till.get(i))
partition_possible = 1;
// Line passes through i, so it neither
// falls left nor right.
if (sum_till.get(i-1) == total_sum - sum_till.get(i))
partition_possible = 1;
}
System.out.printf(partition_possible == 1 ? "YES\n" : "NO\n");
}
// Driven code
public static void main(String[] args)
{
int n = 3;
int x[] = { -1, -2, 1 };
int y[] = { 1, 1, -1 };
int w[] = { 3, 1, 4 };
is_partition_possible(n, x, y, w);
}
}
/* This code contributed by PrinciRaj1992 */ Python3
from collections import defaultdict
# Checking if a plane can be divide by a line
# at 45 degrees such that weight sum is equal
def is_partition_possible(n, x, y, w):
weight_at_x = defaultdict(int)
max_x = -2e3
min_x = 2e3
# Rotating each point by 45 degrees and
# calculating prefix sum.
# Also, finding maximum and minimum x
# coordinates
for i in range(n):
new_x = x[i] - y[i]
max_x = max(max_x, new_x)
min_x = min(min_x, new_x)
# storing weight sum upto x - y point
weight_at_x[new_x] += w[i]
sum_till = []
sum_till.append(0)
# Finding prefix sum
for x in range(min_x, max_x + 1):
sum_till.append(sum_till[-1] +
weight_at_x[x])
total_sum = sum_till[-1]
partition_possible = False
for i in range(1, len(sum_till)):
if (sum_till[i] == total_sum - sum_till[i]):
partition_possible = True
# Line passes through i, so it neither
# falls left nor right.
if (sum_till[i - 1] == total_sum - sum_till[i]):
partition_possible = True
if partition_possible:
print("YES")
else:
print("NO")
# Driven Program
if __name__ == "__main__":
n = 3
x = [-1, -2, 1]
y = [1, 1, -1]
w = [3, 1, 4]
is_partition_possible(n, x, y, w)
# This code is contributed by chitranayal.C#
// Checking if a plane can be divide by a line
// at 45 degrees such that weight sum is equal
using System;
using System.Collections.Generic;
public class GFG{
static void is_partition_possible(int n, int[] x, int[] y, int[] w)
{
Dictionary weight_at_x = new Dictionary();
int max_x = (int) -2e3, min_x = (int) 2e3;
// Rotating each point by 45 degrees and
// calculating prefix sum.
// Also, finding maximum and minimum x
// coordinates
for (int i = 0; i < n; i++)
{
int new_x = x[i] - y[i];
max_x = Math.Max(max_x, new_x);
min_x = Math.Min(min_x, new_x);
// storing weight sum upto x - y point
if(weight_at_x.ContainsKey(new_x))
{
weight_at_x[new_x]+=w[i];
}
else
{
weight_at_x.Add(new_x,w[i]);
}
// weight_at_x[new_x] += w[i];
}
List sum_till = new List();
sum_till.Add(0);
// Finding prefix sum
for (int s = min_x; s <= max_x; s++)
{
if(!weight_at_x.ContainsKey(s))
{
sum_till.Add(sum_till[sum_till.Count - 1]);
}
else
{
sum_till.Add(sum_till[sum_till.Count-1] + weight_at_x[s]);
}
}
int total_sum = sum_till[sum_till.Count-1];
int partition_possible = 0;
for (int i = 1; i < sum_till.Count; i++)
{
if (sum_till[i] == total_sum - sum_till[i])
partition_possible = 1;
// Line passes through i, so it neither
// falls left nor right.
if (sum_till[i-1] == total_sum - sum_till[i])
partition_possible = 1;
}
Console.WriteLine(partition_possible == 1 ? "YES" : "NO");
}
// Driven code
static public void Main (){
int n = 3;
int[] x = { -1, -2, 1 };
int[] y = { 1, 1, -1 };
int[] w = { 3, 1, 4 };
is_partition_possible(n, x, y, w);
}
}
// This code is contributed by rag2127 输出
Yes