1053 path of equal weight

Given a non-empty tree with root R, and with weight W~i~ assigned to each tree node T~i~. 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.the figure

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 < 2^30^, the given weight number. The next line contains N positive numbers where W~i~ (&lt1000) corresponds to the tree node T~i~. 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 {A~1~, A~2~, ..., A~n~} is said to be greater than sequence {B~1~, B~2~, ..., B~m~} if there exists 1 <= k < min{n, m} such that A~i~ = B~i~ for i=1, ... k, and A~k+1~ > B~k+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

题意： 给定一棵树，遍历找到从根节点到叶子节点的一条路径，满足路径的权重和为给定的值。

分析： dfs 遍历树，记录路径。

代码

using namespace std;

int n,m,k;
struct {

    int val;
    vector<int> child;
};
vector< Node >tree;
vector<int> path;

void dfs(int id,int sum)
{
    if(sum>k)
        return;
    if(sum==k)
    {
        if(tree[id].child.size()!=0)
            return;
        for(int i=0;i<path.size()-1;i++)
        {
            printf("%d ",tree[path[i]].val);
        }
        printf("%dn",tree[path[path.size()-1]].val);
        return;
    }
    for(int i=0;i<tree[id].child.size();i++)
    {
        int x=tree[id].child[i];
        path.push_back(x);
        dfs(x,sum+tree[x].val);
        path.pop_back();
    }
}
bool cmp(int a,int b)
{
    return tree[a].val>tree[b].val;
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        tree.resize(n);
        for(int i=0;i<n;i++)
            scanf("%d",&tree[i].val);
        int a,b,c;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            for(int j=0;j<b;j++)
            {
                scanf("%d",&c);
                tree[a].child.push_back(c);
            }
            sort(tree[a].child.begin(),tree[a].child.end(),cmp);
        }
        path.push_back(0);
        dfs(0,tree[0].val);
    }
    return 0;
}