A1053. Path of Equal Weight (30)

Given a non-empty tree with root R, and with weight W_i assigned to each tree node T_i. Theweight 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 begreater 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

因为这个题不只是涉及节点数，还有权重，要用struct，而且要注意，写DFS时可以简便一些，原有的depth可以用sum和替换，只要sum超过s就不需要遍历了，这一点比depth减少遍历很多元素。

因为输出是按权重大小排的，所以莫不如在输入后将结点的顺序按权重大下排一下，这样输出也是按大小顺序输出。给我的启发是，DFS排序市按过程的，与存储的顺序没有关系，所以存储的可以大胆改动，只要父子关系保持，和输入顺序没有关系。

代码如下：

#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>#include<vector>#include<cmath>using namespace std;const int maxn=110;int n,m,s;struct node{int weight;vector<int> child;}node[maxn];bool cmp(int a,int b){return node[a].weight>node[b].weight;}int num=0;int path[maxn]={0};//记录路径void DFS(int index,int numNode,int sum){//index,当前访问结点，numNode,当前path路径上结点个数，sum当前结点权和。if(sum>s) return;if(sum==s){if(node[index].child.size()!=0)return; //权值够，没到叶子点，返回for(int j=0;j<numNode;j++){printf("%d",node[path[j]].weight);if(j!=numNode-1)printf(" ");}printf("\n");return;}for(int i=0;i<node[index].child.size();i++){int child=node[index].child[i];path[numNode]=child;DFS(node[index].child[i],numNode+1,sum+node[node[index].child[i]].weight);}}int main(){//freopen("data.in","r",stdin);scanf("%d %d %d",&n,&m,&s);for(int i=0;i<n;i++)scanf("%d",&node[i].weight);int id,k,temp;for(int j=0;j<m;j++){scanf("%d %d",&id,&k);for(int l=0;l<k;l++){scanf("%d",&temp);node[id].child.push_back(temp);}sort(node[id].child.begin(),node[id].child.end(),cmp);}path[0]=0;DFS(0,1,node[0].weight);return 0;}