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. Figure 1 输入描述: 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. 输出描述: 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. 输入例子: 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 输出例子: 10 5 2 7 10 4 10 10 3 3 6 2 10 3 3 6 2
来自鄙人博客http://blog.csdn.net/sinat_29278271
今天被一道PAT顶级的题目折腾哭了，还好总算是做出来了，决定睡前还是发一篇博客吧，好久没写题解了，顶级的那道我明天再扯。
这道题目其实水的很，要是我考甲级的时候30分的题目是这个，我做梦都会笑出来的。要说算法，其实根本没有算法，从顶点开始暴搜就好了，无需剪枝一路搜到叶子，如果到叶子的时候发现权重与题目要求的相符，就输出路径，为了维护字典序对每个节点的孩子进行一次排序，这样搜索的路径自动遵循字典序从小到大的原则。，然后，然后就没有什么了。
# include
# include
# include
using namespace std;
struct Node
{
int weight;
vector
int child;
Node():child(0){}
bool operator < (const Node& cmper) const
{
return weight > cmper.weight;
}
};
Node v[101];
bool cmper(int a,int b)
{
return v[a] < v[b];
}
void PrintPath(vector
{
vector
for (it = path.begin();it!=path.end();it++)
printf("%d%c",*it,it+1==path.end()?'\n':' ');
}
void dfs(int loca,int rest,vector
{//cout << "loca = " << loca << endl;
Node& cur = v[loca];
path.push_back(cur.weight);
rest -= cur.weight;
if (rest==0&&cur.child==0)
PrintPath(path);
else
{
vector
for (it = cur.edge.begin();it!=cur.edge.end();it++)
dfs(*it,rest,path);
}
path.pop_back();
}
int main()
{
int i,j,k,root,temp;
int n,m,s;
cin >> n >> m >> s;
for (i=0;i cin >> v[i].weight; for (i=0;i { cin >> root >> k; while (k--) { cin >> temp; v[root].edge.push_back(temp); } v[root].child = k; } for (i=0;i sort(v[i].edge.begin(),v[i].edge.end(),cmper); vector dfs(0,s,path); return 0; }
