Codeforces 780C Andryusha and Colored Balloons 搜索dfs

探讨了如何在满足特定条件的情况下,为公园内的节点分配最少数量的不同颜色的气球。通过合理的算法设计,实现了用尽可能少的颜色装饰公园路径。

C. Andryusha and Colored Balloons
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Andryusha goes through a park each day. The squares and paths between them look boring to Andryusha, so he decided to decorate them.

The park consists of n squares connected with (n - 1) bidirectional paths in such a way that any square is reachable from any other using these paths. Andryusha decided to hang a colored balloon at each of the squares. The baloons' colors are described by positive integers, starting from 1. In order to make the park varicolored, Andryusha wants to choose the colors in a special way. More precisely, he wants to use such colors that if ab and c are distinct squares that a and b have a direct path between them, and b and c have a direct path between them, then balloon colors on these three squares are distinct.

Andryusha wants to use as little different colors as possible. Help him to choose the colors!

Input

The first line contains single integer n (3 ≤ n ≤ 2·105) — the number of squares in the park.

Each of the next (n - 1) lines contains two integers x and y (1 ≤ x, y ≤ n) — the indices of two squares directly connected by a path.

It is guaranteed that any square is reachable from any other using the paths.

Output

In the first line print single integer k — the minimum number of colors Andryusha has to use.

In the second line print n integers, the i-th of them should be equal to the balloon color on the i-th square. Each of these numbers should be within range from 1 to k.

Examples
input
3
2 3
1 3
output
3
1 3 2 
input
5
2 3
5 3
4 3
1 3
output
5
1 3 2 5 4 
input
5
2 1
3 2
4 3
5 4
output
3
1 2 3 1 2 
Note

In the first sample the park consists of three squares: 1 → 3 → 2. Thus, the balloon colors have to be distinct.

Illustration for the first sample.

In the second example there are following triples of consequently connected squares:

  • 1 → 3 → 2
  • 1 → 3 → 4
  • 1 → 3 → 5
  • 2 → 3 → 4
  • 2 → 3 → 5
  • 4 → 3 → 5
We can see that each pair of squares is encountered in some triple, so all colors have to be distinct.
Illustration for the second sample.

In the third example there are following triples:

  • 1 → 2 → 3
  • 2 → 3 → 4
  • 3 → 4 → 5
We can see that one or two colors is not enough, but there is an answer that uses three colors only.
Illustration for the third sample.


搜索写的不熟练,一看到树就害怕,抽自己一巴掌

扪心自问。。。。你都在干什么鬼事情


#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
using namespace std;
const int maxn = 2e5+10;
const int inf = 0x3f3f3f3f;
vector<int> save[maxn];
int color[maxn];
int n,res;
void dfs(int now, int father){
    int i,k;
    k = 0;
    for(i=0;i<save[now].size();i++){
        if(save[now][i]==now||color[save[now][i]]!=inf){
            continue;
        }
        k++;
        while(color[now]==k||color[father]==k){
            k++;
        }
        color[save[now][i]] = k;
        dfs(save[now][i], now);
    }
    res = res<k?k:res;
}
int main(){
    int i,a,b;
    scanf("%d",&n);
    for(i=1;i<=n-1;i++){
        scanf("%d%d",&a,&b);
        save[a].push_back(b);
        save[b].push_back(a);
    }
    memset(color, inf, sizeof(color));
    color[0] = 0;
    color[1] = 1;
    dfs(1, 0);
    printf("%d\n",res);
    i = 1;
    while(color[i]!=inf){
        printf("%d ",color[i++]);
    }
    return 0;
}








### 解题思路 #### 问题描述 Codeforces 1678C - Tokitsukaze and Strange Inequality 是一道关于排列组合与前缀和的应用问题。给定一个长度为 \( n \) 的排列数组 \( p \),需要统计满足条件 \( a < b < c < d \) 并且 \( p_a < p_c \) 同时 \( p_b > p_d \) 的四元组数量。 --- #### 核心思想 由于数据规模较小 (\( n \leq 5000 \)),可以直接通过枚举的方式解决问题。为了降低时间复杂度,引入 **前缀和** 技术来加速计算过程[^3]。 具体来说: - 枚举变量 \( a \) 和 \( c \),固定它们之后,目标是快速找到符合条件的 \( b \) 和 \( d \)。 - 使用预处理好的前缀和数组 `num` 来高效查询某个范围内满足特定关系的数量。 - 定义辅助数组 `sum` 表示对于固定的区间范围内的某些约束条件下的累积计数结果。 --- #### 实现细节 ##### 步骤一:构建前缀和数组 `num` 定义二维数组 `num[i][j]`,其中 `num[i][j]` 表示在序列的前 \( i \) 项中,有多少个元素大于 \( j \)。 该数组可以通过如下方式初始化: ```python n = len(p) max_val = max(p) # 初始化 num 数组 num = [[0] * (max_val + 2) for _ in range(n + 1)] for i in range(1, n + 1): for j in range(max_val + 1, -1, -1): # 反向遍历以保持正确性 if p[i - 1] > j: num[i][j] = num[i - 1][j] + 1 else: num[i][j] = num[i - 1][j] ``` 上述代码的时间复杂度为 \( O(n \cdot m) \),其中 \( m \) 是数组中的最大值。 --- ##### 步骤二:定义并填充辅助数组 `sum` 定义另一个二维数组 `sum[i][j]`,它表示当 \( a=i \), \( c=j \) 时,在区间 \([a+1, c-1]\) 中满足 \( p[b] > p[d] \) 的总贡献次数。 利用动态规划的思想逐步更新此数组: ```python sum_ = [[0] * (n + 1) for _ in range(n + 1)] bucket = [0] * (max_val + 1) for l in range(n - 1, 0, -1): bucket[p[l]] += 1 for r in range(l + 2, n + 1): sum_[l][r] = sum_[l][r - 1] + (num[r - 1][p[r - 1]] - num[l][p[r - 1]]) ``` 这里的关键在于如何有效累加当前区间的合法贡献,并借助之前已经计算的结果减少重复运算。 --- ##### 步骤三:枚举所有可能的 \( a \) 和 \( c \) 最后一步是对所有的 \( a \) 和 \( c \) 进行双重循环,并将对应位置上的 `sum[a][c]` 加入最终答案中: ```python result = 0 for a in range(1, n - 2): for c in range(a + 2, n): result += sum_[a][c] print(result) ``` 整个算法的核心部分即完成以上三个阶段的操作即可实现高效的解决方案。 --- ### 总结 本题主要考察的是对多重嵌套结构的有效简化以及合理运用前缀和技巧的能力。通过巧妙设计的数据结构能够显著提升程序运行效率至可接受水平。
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