1053. Path of Equal Weight (30)

题目连接：https://www.patest.cn/contests/pat-a-practise/1053原题如下：

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 < 2³⁰, 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., 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

这道是我觉得是一道很典型的DFS题。我自己递归水平不行，参考了陈小旭的解法。值得多次回看！！！

#include<stdio.h>
#include<vector>
#include<algorithm>
#define MAXN 105
using namespace std;

typedef struct Node
{
    int weight;
    vector<int>son;
}node;
node bur[MAXN];
int S;
vector<int>ans;
vector<vector<int> >ans_V;

void DFS(int p,int sum)
{
    sum+=bur[p].weight;
    ans.push_back(bur[p].weight);
    if (sum<S)
    {
        for (int i=0;i<bur[p].son.size();i++)DFS(bur[p].son[i],sum);
        ans.pop_back();  //没有孩子，但仍然小于S
        return ;
    }
    else if (sum>S)
    {
        ans.pop_back();
        return ;
    }
    else if (sum==S)
    {
        if (bur[p].son.size()==0)
        {
            ans_V.push_back(ans);
            ans.pop_back();
        }
        else
        {
            ans.pop_back();
        }
        return ;
    }
}

int cmp(vector<int>a,vector<int>b)
{
    int len1,len2;
    len1=a.size();
    len2=b.size();

    int i;
    for (i=0;i<len1 && i<len2;i++)
    {
        if (a[i]==b[i])continue;
        else  return (a[i]>b[i]);
    }
    return (len1>len2);
}

int main()
{
    int N,M;
    scanf("%d %d %d",&N,&M,&S);
    for (int i=0;i<N;i++)scanf("%d",&bur[i].weight);
    int p,k,w;
    while(M--)
    {
        scanf("%d %d",&p,&k);
        while(k--)
        {
            scanf("%d",&w);
            bur[p].son.push_back(w);
        }
    }
    DFS(0,0);
    sort(ans_V.begin(),ans_V.end(),cmp);
    for (int i=0;i<ans_V.size();i++)
    {
        for (int j=0;j<ans_V[i].size();j++)
        {
            if (j==0)printf("%d",ans_V[i][j]);
            else printf(" %d",ans_V[i][j]);
        }
        printf("
");
    }
    return 0;
}