（pat）A1053 Path of Equal Weight

Given a non-empty tree with root R, and with weight Wi assigned to each tree node Ti. 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 Wi (<1000) corresponds to the tree node Ti. 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, …, An} is said to be greater than sequence {B1, B2, …, Bm} if there exists 1 <= k < min{n, m} such that Ai = Bi for i=1, … k, and Ak+1 > Bk+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

对于勉强算是一个图论选手还是太简单啦

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
vector<int> graph[105];
int val[105];
int n ,m,s;
int a[105];
int countt=0;
int sum=0;
void printt()
{
    for(int i=0;i<countt-1;i++)
        printf("%d ",a[i]);
    printf("%d\n",a[countt-1]);
}
void dfs(int v,int fa)
{
    a[countt++]=val[v];
    sum+=val[v];
    if(sum>=s)
    {
        if(sum==s&&graph[v].size()==0)
            printt();
        sum-=val[v];
        countt--;
        return;
    }
    else
    {
        for(int i=0;i<graph[v].size();i++)
        {
            int u=graph[v][i];
            if(v==u)
                continue;
            dfs(u,v);
        }
        sum-=val[v];
        countt--;
    }

}
bool cmp(int x,int y)
{
    return val[x]>val[y];
}
int main()
{
    scanf("%d%d%d",&n,&m,&s);
    for(int i=0;i<n;i++)
        scanf("%d",val+i);
    int x,num,y;
    while(m--)
    {
        scanf("%d%d",&x,&num);
        while(num--)
        {
            scanf("%d",&y);
            graph[x].push_back(y);
        }
    }
    for(int i=0;i<n;i++)
        sort(graph[i].begin(),graph[i].end(),cmp);//字典序
    dfs(0,-1);
    return 0;
}