PAT [A1053]-Path of Equal Weight

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.

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 1:
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 1:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

解题思路：
按weight大小构造二叉树，并用二叉树来表示N叉树，然后利用深度搜索来得到结果。

AC代码：

#include <cstdio>const int maxn = 110;int N, M, S;int dat[maxn];struct Node{    int value;    int id;    Node* lChild;    Node* nextChild;}node[maxn];void insert(Node* root, int id){    if (root->lChild == NULL){        root->lChild = &node[id];    }    else{        Node* tmp;        if (node[id].value > root->lChild->value) {            tmp = root->lChild;            root->lChild = &node[id];            root = root->lChild;        }        else {            root = root->lChild;            tmp = &node[id];        }        while (root->nextChild != NULL) {            if (tmp->value > root->nextChild->value){                Node* ttmp = root->nextChild;                root->nextChild = tmp;                tmp = ttmp;                root = root->nextChild;            }            else root = root->nextChild;        }        root->nextChild = tmp;    }}void DFS(int sum, int cnt, int id){    Node tmp = node[id];    sum += tmp.value;    //printf("%d %d\n", sum, id);    if (sum > S) return;    else{        dat[cnt++] = tmp.value;        if (sum == S){            if (tmp.lChild != NULL) return;            for (int i = 0; i < cnt; i++){                printf("%d", dat[i]);                if (i < cnt - 1) printf(" ");            }            printf("\n");            return;        }        else{            if (tmp.lChild == NULL) return;            tmp = node[tmp.lChild->id];            DFS(sum, cnt, tmp.id);            while (tmp.nextChild != NULL){                DFS(sum, cnt, tmp.nextChild->id);                tmp = node[tmp.nextChild->id];            }            return;        }    }}int main(){    freopen("C:\\Users\\Administrator\\Desktop\\test.txt", "r", stdin);    while (scanf("%d%d%d", &N, &M, &S) != EOF){        for (int i = 0; i < N; i++){            scanf("%d", &node[i].value);            node[i].lChild = node[i].nextChild = NULL;            node[i].id = i;            dat[i] = 0;        }        int id, num;        for (int i = 0; i < M; i++){            int tmpID;            scanf("%d %d", &id, &num);            for (int i = 0; i < num; i++){                scanf("%d", &tmpID);                insert(&node[id], tmpID);            }        }        DFS(0, 0, 0);    }    fclose(stdin);    return 0;}