Zero Sum(DFS)

本文介绍了一种寻找特定长度数列中能构成零和表达式的算法实现。通过对每个数位间插入加号、减号或留空的操作,程序能够找出所有可能组合,并输出那些求和为零的表达式。

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Problem Description:


         Consider the sequence of digits from 1 through N (where N=9) in increasing order: 1 2 3 ... N. 


Now insert either a '+' for addition or a '-' for subtraction or a ' ' [blank] to run the digits together between each pair of digits (not in front of the first digit). Calculate the result that of the expression and see if you get zero. 


Write a program that will find all sequences of length N that produce a zero sum. 


PROGRAM NAME: zerosum
INPUT FORMAT
A single line with the integer N (3 <= N <= 9). 
Input:


The input includes several cases. For each case, input a single line with the integer N (3 <= N <= 9). 


Output:


For each case, the output format is:
In ASCII order, show each sequence that can create 0 sum with a '+', '-', or ' ' between each pair of numbers. 
Sample Input:


7
Sample Output:


1+2-3+4-5-6+7
1+2-3-4+5+6-7
1-2 3+4+5+6+7
1-2 3-4 5+6 7
1-2+3+4-5+6-7
1-2-3-4-5+6+7
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

#define INF 0x3f3f3f3f

int num[20];
int symbol[20];
int n;

int cou()
{
    int i,j;
    int sum=0;
    int ca;
    for(i=0;i<=10;i++)
        num[i]=i;
    for(i=1;i<n;i++)
    {
        if(symbol[i]==0)
        {
            num[i+1]+=10*num[i];
            num[i]=0;
        }
    }
    /*for(i=1;i<=n;i++)
        printf("%d ",num[i]);*/
    //printf("\n");
    sum=0;
    ca=1;
    for(i=1;i<n;i++)
    {
        if(symbol[i]==0)
            continue;
        if(symbol[i]==1)
        {
            if(ca==1)
            {
                sum+=num[i];
            }
            else
                sum-=num[i];
            ca=1;
        }
        if(symbol[i]==2)
        {
            if(ca==1)
            {
                sum+=num[i];
            }
            else
                sum-=num[i];
            ca=2;
        }
    }
    if(ca==1)
    {
        sum+=num[n];
    }
    else
        sum-=num[n];
    if(sum)
        return;
    for(i=1;i<n;i++)
    {
        printf("%d",i);
        if(symbol[i]==1)
            printf("+");
        else if(symbol[i]==2)
            printf("-");
        else
            printf(" ");
    }
    printf("%d\n",n);
}


void dfs(int a)
{
    int i;
    if(a==n)
    {
        cou();
        return;
    }
    for(i=0;i<3;i++)
    {
        symbol[a]=i;
        dfs(a+1);
    }
}

int main()
{
    int i,j,k;
    while(scanf("%d",&n)==1)
    {
        dfs(1);
    }
    return 0;
}

### 使用DFS解决树染色问题的最少时间计算 #### 1. 树染色问题背景 树染色问题的目标是在满足特定约束条件下,为树上的节点分配颜色。对于本问题而言,假设存在多种颜色可供选择,并且希望找到一种最优解使得总成本最低或所需操作次数最少。 在实际应用中,“最少时间”的定义可以根据具体需求有所不同。例如: - 如果每改变一次颜色需要花费单位时间,则目标是最小化总的变色次数。 - 或者考虑更复杂的权重模型,在这种情况下需综合评估各路径长度等因素影响。 此处讨论的是基于简单规则下的最小时间计算方式——即假定切换任意两不同颜色均耗费固定代价\(T\)[^1]。 --- #### 2. 解决方案设计 ##### 数据结构准备 利用邻接表存储无向树的信息以便于后续快速查询连接关系。同时引入辅助数组用于追踪当前状态下各个顶点所处的颜色以及访问状态等信息。 ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = 1e5 + 7; vector<vector<int>> tree(MAXN); // Tree represented as adjacency list. int color_time[MAXN]; // Minimum time to reach this state at each node. bool visited_node[MAXN]; int total_min_time = 0; // Global variable storing final answer. void initialize_data(int nodes){ fill(color_time, color_time + nodes + 1, INT_MAX); fill(visited_node, visited_node + nodes + 1, false); } // Function performing depth-first search while calculating minimum times. void dfs_calculate_minimum_times(int current_vertex, int previous_color, int accumulated_time){ if (visited_node[current_vertex]){ return ; } visited_node[current_vertex] = true; // Calculate new color based on some logic here... // For simplicity assume we toggle between two colors only. int next_color = (previous_color == 1 ? 2 : 1); // Transition cost from last vertex&#39;s color change plus existing accumulation. int transition_cost = ((next_color != previous_color)? 1 : 0 ); int updated_accumulated_time = accumulated_time + transition_cost; // Update the best known result for reaching this point with certain configuration. if(updated_accumulated_time < color_time[current_vertex]){ color_time[current_vertex] = updated_accumulated_time; // Propagate further into children vertices recursively now that local optimum established. for(auto child : tree[current_vertex]){ if(child != parent_of_current_in_dfs_traversal){ // Avoid going back up immediately after coming down an edge. dfs_calculate_minimum_times(child, next_color, updated_accumulated_time); } } } } ``` --- #### 3. 主函数调用流程 首先读取输入数据构建完整的树形拓扑结构,接着指定任一叶节点作为起点启动递归过程直至覆盖全图范围内的每一个角落位置为止最后统计汇总所得全局最短耗时时长数值即可得出结论[^2]. ```cpp int main(){ ios::sync_with_stdio(false); cin.tie(NULL); int num_nodes, edges_count; cin >> num_nodes >> edges_count; for(int i=0;i<edges_count;++i){ int u,v; cin>>u>>v; tree[u].emplace_back(v); tree[v].emplace_back(u); } initialize_data(num_nodes); // Start traversal arbitrarily picking first unprocessed leaf or root-like entity within structure formed by provided inputs. dfs_calculate_minimum_times(1 /*starting_point*/, 0/*no prior color initially*/, 0/*zero initial timing*/); long long overall_sum=accumulate(color_time+1,color_time+(num_nodes)+1,(long long )0); cout<<overall_sum<<"\n"; return 0; } ``` --- #### 4. 复杂度分析 该算法主要依赖单次遍历整棵树的操作模式完成任务因此总体性能表现良好属于线性级别范畴内\[O(N)\],其中\(N\)代表参与运算的所有节点数目之和[^3]。 --- ###
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