1053 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 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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where W​i​​ (<) 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 { is said to be greater than sequence { if there exists 1 such that A​i​​=B​i​​ for ,, 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<string>
#include<algorithm>
using namespace std;
const int maxn=100;
struct node{
    int id;
    int value;
    int num1=0;
    int sub[maxn];
};
node str[maxn];
int a[maxn];
int n,m,w,flag=0;
string s[maxn];
int cmp(string s1,string s2)
{
    return s1>s2;
}
void dfs(int id1,int w1,int cnt)
{
    if(w1<0)
    return ;
    if(w1==0&&str[id1].num1==0)
    {
        s[flag]="";
    for(int i=0;i<cnt;i++)
    {
        s[flag]=s[flag]+char(a[i]+'0');
    }
    //cout<<flag<<endl;
    //cout<<s[flag]<<endl;
    flag++;
   }
    for(int i=0;i<str[id1].num1;i++)
    {
        a[cnt]=str[str[id1].sub[i]].value;
        dfs(str[id1].sub[i],w1-str[str[id1].sub[i]].value,cnt+1);
    }
}
int main()
{
    int k,k1,temp;
    cin>>n>>m>>w;
    for(int i=0;i<n;i++)
    {
        cin>>str[i].value;
    }
    while(m--)
    {
    cin>>k>>k1;
    str[k].num1=k1;
    for(int i=0;i<k1;i++)
    {
        cin>>temp;
        str[k].sub[i]=temp;
    }
    }
    a[0]=str[0].value;
    dfs(0,w-str[0].value,1);
    sort(s,s+flag,cmp);
    //cout<<flag<<endl;
    for(int i=0;i<flag;i++)
    {
        for(int j=0;j<s[i].size();j++)
        {
            if(j==0)
            cout<<int(s[i][j]-'0');
            else
            cout<<" "<<int(s[i][j]-'0');
        }
        cout<<endl;
    }
    return 0;
 } 

总有一些case过不了