1053. Path of Equal Weight (30)

 

时间限制

10 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 Figure 1.

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 W_i (<1000) 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., 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 2410 2 4 3 5 10 2 18 9 7 2 2 1 3 12 1 8 6 2 200 4 01 02 03 0402 1 0504 2 06 0703 3 11 12 1306 1 0907 2 08 1016 1 1513 3 14 16 1717 2 18 19

Sample Output:

10 5 2 710 4 1010 3 3 6 210 3 3 6 2

 

 

#include <stdio.h>#include <stdlib.h>#define MAX 110typedef struct node{int weight;int tail;int children[MAX];}NODE;NODE tree[MAX];int solution[MAX],SolutionTail;int N, M, target;int cmp(const void *a, const void *b){return tree[*(int*)b].weight - tree[*(int*)a].weight;}void DFS(int node, int CurrentWeight){int i;CurrentWeight = CurrentWeight + tree[node].weight;if (tree[node].tail == 0)  // reach a leaf{if (CurrentWeight == target){for (i = 0; i < SolutionTail-1; i++){printf("%d ", solution[i]);}printf("%d\n", solution[i]);}return;}else {if (CurrentWeight > target){return;}else{qsort(tree[node].children, tree[node].tail, sizeof(tree[node].children[0]), cmp);for (i = 0; i < tree[node].tail; i++){solution[SolutionTail++] = tree[tree[node].children[i]].weight;DFS(tree[node].children[i], CurrentWeight);SolutionTail--;}}}}int main(){int id, k;int CurrentWeight;//freopen("d://input.txt", "r", stdin);scanf("%d%d%d", &N, &M, &target);for (int i = 0; i < N; i++){scanf("%d", &tree[i].weight);}for (int i = 0; i < M; i++){scanf("%d%d", &id, &k);for (int j = 0; j < k; j++){scanf("%d", &tree[id].children[tree[id].tail++]);}}solution[SolutionTail++] = tree[0].weight;CurrentWeight = 0;DFS(0, CurrentWeight);return 0;}