Path of Equal Weight

1053. Path of Equal Weight (30)

时间限制

100 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_ifor 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

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提交代码

#include<stdio.h>#include<string.h>#include<vector>#include<algorithm>using namespace std;const int maxn=110;struct node{int weight;vector<int> child;}Node[maxn];bool cmp(int a,int b){return Node[a].weight>Node[b].weight;}int n,m,s;int path[maxn];void DFS(int index,int numNode,int sum){if(sum>s) return ;if(sum==s){if(Node[index].child.size()!=0) return;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++){int child=Node[index].child[i];path[numNode]=child;DFS(child,numNode+1,sum+Node[child].weight);} }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);}sort(Node[id].child.begin(),Node[id].child.end(),cmp);}path[0]=0;DFS(0,1,Node[0].weight);return 0;}