A1053. 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 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 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

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef struct NODE{
    int weigh;
    vector<int> child;
}node;
node tree[101];
int N, M, S;
vector<int> ans;
bool cmp(int a, int b){
    return tree[a].weigh > tree[b].weigh;
}
void firstRoot(int root, int sum){
    sum += tree[root].weigh;
    ans.push_back(root);
    if(tree[root].child.size() == 0 && sum == S){
        int len = ans.size();
        for(int i = 0; i < len; i++){
            if(i == len - 1)
                printf("%d
", tree[ans[i]].weigh);
            else printf("%d ", tree[ans[i]].weigh);
        }
        ans.pop_back();
        return;
    }
    if(sum > S || tree[root].child.size() == 0){
        ans.pop_back();
        return;
    }
    if(sum < S){
        int childnum = tree[root].child.size();
        for(int i = 0; i < childnum; i++)
            firstRoot(tree[root].child[i], sum);
    }
    ans.pop_back();
}
int main(){
    scanf("%d%d%d", &N, &M, &S);
    for(int i = 0; i < N; i++)
        scanf("%d", &tree[i].weigh);
    int index, childn, tempc;
    for(int i = 0; i < M; i++){
        scanf("%d%d", &index, &childn);
        for(int j = 0; j < childn; j++){
            scanf("%d", &tempc);
            tree[index].child.push_back(tempc);
        }
        sort(tree[index].child.begin(), tree[index].child.end(), cmp);
    }
    firstRoot(0, 0);
    cin >> N;
    return 0;
}

总结：

1、题意：给出一颗节点带权的树，给出指定权的和S，找出从根到叶节点的权的和为S的所有路径，如果有多条，则按照权的大小输出。

2、在写先根遍历的时候不要忘记push和pop，当在条件语句外加入当前节点时，需要在每个条件语句内部都pop，在整个函数结尾处也pop。 当找到符合条件的解时，不是将ans清空，而是仍然仅仅pop当前选择的元素。

3、要求从大到小输出，可以在读完每个节点时，对其孩子节点进行排序，权值大的孩子排在前面即可。