1053. Path of Equal Weight (30)

我就是我 2022-05-31 13:39 149阅读 0赞

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. The **weight of a path from R to L** is defined to be the sum of the weights of all the nodes along the path from R to any leaf node L.

Now given any weighted tree, you are supposed to find all the paths with their weights equal to a given number. For example, let's consider the tree showed in Figure 1: for each node, the upper number is the node ID which is a two-digit number, and the lower number is the weight of that node. Suppose that the given number is 24, then there exists 4 different paths which have the same given weight: \{10 5 2 7\}, \{10 4 10\}, \{10 3 3 6 2\} and \{10 3 3 6 2\}, which correspond to the red edges in Figure 1.

![5b_mj58f84v11w.jpg][]  
Figure 1

**Input Specification:**

Each input file contains one test case. Each case starts with a line containing 0 < N <= 100, the number of nodes in a tree, M (< N), the number of non-leaf nodes, and 0 < S < 230, the given weight number. The next line contains N positive numbers where Wi (<1000) corresponds to the tree node Ti. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 00.

**Output Specification:**

For each test case, print all the paths with weight S in **non-increasing** order. Each path occupies a line with printed weights from the root to the leaf in order. All the numbers must be separated by a space with no extra space at the end of the line.

Note: sequence \{A1, A2, ..., An\} is said to be **greater than** sequence \{B1, B2, ..., Bm\} if there exists 1 <= k < min\{n, m\} such that Ai = Bi for i=1, ... k, and Ak+1 > Bk+1.

**Sample Input:**

20 9 24
    10 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 2
    00 4 01 02 03 04
    02 1 05
    04 2 06 07
    03 3 11 12 13
    06 1 09
    07 2 08 10
    16 1 15
    13 3 14 16 17
    17 2 18 19

**Sample Output:**

10 5 2 7
    10 4 10
    10 3 3 6 2
    10 3 3 6 2

--------------------

题目大意：

代码：

#include<stdio.h>
    #include<vector>
    #include<map>
    #include<algorithm>
    using namespace std;
    struct node
    {
        int index;
        int weight;
    };
    map<int,vector<struct node> > Map;
    int weight[101];
    int sum,n,m,k;
    vector<int> path;
    bool cmp(struct node a,struct node b)
    {
        return a.weight>b.weight;
    }
    void DFS(int root)
    {
        if(sum<k)
        {
            sort(Map[root].begin(),Map[root].end(),cmp);
            for(int i=0;i<Map[root].size();i++)
            {
                sum+=Map[root][i].weight;
                path.push_back(Map[root][i].index);
                DFS(Map[root][i].index);
                sum-=Map[root][i].weight;
                path.pop_back();
            }
        }
        else if(sum==k&&Map[root].size()==0)
        {
            for(int i=0;i<path.size();i++)
            {
                if(i==0)
                {
                    printf("%d",weight[path[i]]);
                }
                else
                {
                    printf(" %d",weight[path[i]]);
                }
            }
            printf("\n");
        }
    }
    int main()
    {
        int i,j,t,index,l;
        scanf("%d %d %d",&n,&m,&k);
        for(i=0;i<n;i++)
        {
            scanf("%d",&weight[i]);
        }
        for(i=0;i<m;i++)
        {
            scanf("%d",&index);
            scanf("%d",&l);
            for(j=0;j<l;j++)
            {
                scanf("%d",&t);
                struct node tmp;
                tmp.index=t;
                tmp.weight=weight[t];
                Map[index].push_back(tmp);
            }
        }
        sum=weight[0];
        path.push_back(0);
        DFS(0);
        return 0;
    }

[5b_mj58f84v11w.jpg]: http://nos.patest.cn/5b_mj58f84v11w.jpg