1053 Path of Equal Weight (30 分)

题目描述

Given a non-empty tree with root R, and with weight W i W_{i} Wi​ assigned to each tree node T ​ i T_{​i} 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.

输入

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< 2 30 2^{30} 230 , the given weight number. The next line contains N positive numbers where W i W_{i} Wi​(<1000) corresponds to the tree node T ​ i T_{​i} 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.

输出

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 { A 1 A_{1} A1​, A 2 A_{2} A2​,⋯, A n A_{n} An​} is said to be greater than sequence { B 1 B_{1} B1​, B 2 B_{2} B2​,⋯, B n B_{n} Bn​} if there exists 1≤k<min{n,m} such that A ​ i A_{​i} A​i​= B i B_{i} Bi​ for i=1,⋯,k, and A ​ k + 1 A_{​k+1} A​k+1​> B k + 1 B_{k+1} Bk+1​.

样例输入

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

样例输出

10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2

实现代码

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

const int MAXN = 110;
int n, m, s;
int path[MAXN];

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

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

// 当前访问结点为index，numNode为当前路径path上的结点个数
// sum为当前的结点权值和
void DFS(int index, int numNode, int sum) {
    if(sum > s) {
        return;
    }
    if(sum == s) {
        if(Node[index].child.size() != 0) {
            return;
        }
        for(int i = 0; i < numNode; i++) {
            printf("%d", Node[path[i]].weight);
            if(i < numNode - 1) {
                printf(" ");
            } else {
                printf("\n");
            }
        }
        return;
    }
    for(int i = 0; i < Node[index].child.size(); i++) {
        int child = Node[index].child[i];
        int sum1 = sum + Node[child].weight;
        if(sum1 > s) {
            continue;
        } else {
            path[numNode] = child;
            DFS(child, numNode + 1, sum1);
        }
    }
}

int main() {
    scanf("%d%d%d", &n, &m, &s);
    for(int i = 0; i < n; i++) {
        scanf("%d", &Node[i].weight);
    }
    int id, k, child;
    for(int i = 0; i < m; i++) {
        scanf("%d%d", &id, &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &child);
            Node[id].child.push_back(child);
        }
        sort(Node[id].child.begin(), Node[id].child.end(), cmp); //按权值从大到小排序
    }
    path[0] = 0;
    DFS(0, 1, Node[0].weight);
    return 0;
}

标签：Node,10,1053,weight,int,Equal,child,Path,path
来源： https://blog.csdn.net/Lerbronjames/article/details/113822853