（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-i
4000
ncreasing 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;
}