1053 Path of Equal Weight （30 分）（树的遍历）

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 the following figure: 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 the figure.

 

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

(<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 {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

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cctype>
#include <string.h>
#include <cstdio>
using namespace std;
int target;
struct NODE{
    int w;
    vector<int> child;
};
vector<NODE> v;
vector<int> path;
void dfs(int index,int nodeNum,int sum){
    if(sum>target) return;
    if(sum==target){
        if(v[index].child.size()!=0) return;
        for(int i=0;i<nodeNum;i++)
            printf("%d%c",v[path[i]].w,i!=nodeNum-1?' ':'\n');
        return;
    }
    for(int i=0;i<v[index].child.size();i++){
        int node=v[index].child[i];
        path[nodeNum]=node;
        dfs(node,nodeNum+1,sum+v[node].w);
    }
}
bool cmp(int a,int b){
    return v[a].w>v[b].w;   //根据权值交换序号
}
int main(){
    int n,m,node,k;
    cin>>n>>m>>target;
    v.resize(n);
    path.resize(n);
    for(int i=0;i<n;i++)
        cin>>v[i].w;
    for(int i=0;i<m;i++){
        cin>>node>>k;
        v[node].child.resize(k);
        for(int j=0;j<k;j++)
            cin>>v[node].child[j];
        sort(v[node].child.begin(),v[node].child.end(),cmp);
    }
    dfs(0,1,v[0].w);
    return 0;
}