PAT A1053

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.

请输入图片描述
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 ID1 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

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#include "stdio.h"
//#include "math.h"
#include "string.h"
//#include "iostream"
//#include "stdlib.h"
#include "vector"
//#include "set"
//#include "map"
//#include "stack"
//#include "queue"
#include "algorithm"
using namespace std;

//typedef long long LL;
const int maxn = 110;
struct node {
int weight;//数据域
vector<int> child;//指针域
}Node[maxn];
bool cmp(int a, int b){
return Node[a].weight > Node[b].weight;
}
int n, m, S;//结点数、边数、给定的和
int path[maxn];//记录路径
//当前访问结点为index, nowNode为当前路径path上的结点个数
//sum为当前的结点weight和
void DFS(int index, int numNode, int sum){
if (sum > S) return;//当前和不等于S
if (sum == S) {//当前和等于S
if (Node[index].child.size()) return;//当前和等于S但不是叶子结点
for (int i = 0; i < numNode; i++) {
printf("%d", Node[path[i]].weight);
if (i < numNode - 1) printf(" ");
else printf("\n");
}
return;
}
for (int i = 0; i < Node[index].child.size(); i++) {//枚举孩子结点
int child = Node[index].child[i];
path[numNode] = child;//将结点child添加到路径path末尾
DFS(child, numNode + 1, sum + Node[child].weight);//递归下一层
}
}
int main(){
scanf("%d%d%d", &n, &m, &S);
for (int i = 0; i < n; i++) {
scanf("%d", &Node[i].weight);
}
int id, k, child;
for (int i = 0; i < m; i++) {
scanf("%d%d", &id, &k);
for (int j = 0; j < k; j++) {
scanf("%d", &child);
Node[id].child.push_back(child);//child为结点id的孩子
}
sort(Node[id].child.begin(), Node[id].child.end(), cmp);
}
path[0] = 0;//路径设置第一个结点为0号结点
DFS(0, 1, Node[0].weight);//DFS求解
return 0;
}