1053 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~. 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 < 2^30^, the given weight number. The next line contains N positive numbers where W~i~ (&lt1000) 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 {A~1~, A~2~, ..., A~n~} is said to be greater than sequence {B~1~, B~2~, ..., 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

作者: CHEN, Yue

单位: PAT联盟

时间限制: 100ms

内存限制: 64MB

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 110;
struct Node{
    int weight;
    vector<int> child;
}node[maxn];
int path[maxn] = {0};
int n,m,s;
bool cmp(int a, int b){
    return node[a].weight > node[b].weight; 
}

void DFS(int index, int numNode, int sum){
    if(sum > s) return;
    if(sum == s){
        if(node[index].child.size() != 0) return;
        else{
            for(int i = 0; i < numNode; i++){
                printf("%d",node[path[i]].weight);
                if(i < numNode - 1) printf(" ");
                else printf("
"); 
            }
        }
    }
    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 father,child,k;
    for(int i = 0; i < m; i++){
        scanf("%d%d",&father,&k);
        for(int j = 0; j < k; j++){
            scanf("%d",&child);
            node[father].child.push_back(child);
        }
        sort(node[father].child.begin(),node[father].child.end(),cmp);
    }
    DFS(0,1,node[0].weight);
    return 0;
}