PAT A1053 Path of Equal Weight +静态树+深搜

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 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<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:

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn=110;

struct node{
    int weight;
    vector<int> child;
}Node[maxn];

int n,m,S;
vector<int> path;

bool cmp(int a,int b){
    return Node[a].weight>Node[b].weight;
}

void printPath(){
    for(int i=0;i<path.size();i++){
        printf("%d",path[i]);
        if(i!=path.size()-1)
            printf(" ");
    }
    printf("\n");
}

//处理下标为id的结点，前面的结点权重和为sum
void DFS(int id,int sum){
    //大了或者（相等但不是叶节点的）咱就不递归了

    //如果加上这个节点正好==S且这个节点是叶节点
    if(sum+Node[id].weight==S && Node[id].child.size()==0){
        path.push_back(Node[id].weight); //一加
        printPath();
        path.pop_back(); //一关
    }else if(sum+Node[id].weight<S){
        sum+=Node[id].weight;  //一加
        path.push_back(Node[id].weight);
        int c;
        for(int i=0;i<Node[id].child.size();i++){ //深搜自己的孩子结点
            c=Node[id].child[i];
            DFS(c,sum);
        }
        path.pop_back(); //一关
    }
}

int main()
{
    cin >> n >> m >> S;
    for(int i=0;i<n;i++){
        scanf("%d",&Node[i].weight);
    }
    int index,cnum,c;
    for(int i=0;i<m;i++){
        scanf("%d%d",&index,&cnum);
        for(int j=0;j<cnum;j++){
            scanf("%d",&c);
            Node[index].child.push_back(c);
        }
        sort(Node[index].child.begin(),Node[index].child.end(),cmp); //按孩子的权重从大到小排序
    }

    DFS(0,0);
    return 0;
}

PAT A1053 Path of Equal Weight +静态树+深搜

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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