A1053 Path of Equal Weight (30 分)
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:
10 5 2 7
10 4 10
10 3 3 6 2
10 3 3 6 2
题意:
输入: 第一行:N个结点,M个非叶节点,权值之和S
第二行:按序给出每个节点的权值
后面M行:给出每个非叶节点的孩子信息;第三行:00节点有4个孩子节点,分别为:01 02 03 04
求所有从根节点出发到叶子结点的路径,使得每条路径上的权值之和为S;如果有多条这样的路径,按路径非递增的顺序输出。
思路:
本题的难点在于牵扯到统计的量比较多,还有输出排序的问题
使用DFS函数,参数为 index, numNode(当前路径上的节点个数,即就是深度),sum(权值之和),递归访问这棵树,即可找出路径。
定义path 数组存储当前的路径
排序问题:在输入时,就对每一个非叶节点的孩子节点进行排序,按照权值从大到小排序,这样最终得到的路径顺序即为题目要求顺序
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 110;
struct node{
int weight;
vector<int> child;
}Node[maxn];
bool cmp(int a, int b){
return Node[a].weight > Node[b].weight;
}
int n, m, S;
int path[maxn];
void DFS(int index, int numNode, int sum){
if(sum > S) //当前和超过S时,直接返回
return;
if(sum == S){
if(Node[index].child.size() != 0)
return; //还没到叶子结点时,直接返回
for(int i = 0; i < numNode; i++){
printf("%d", Node[path[i]].weight);
if(i < numNode - 1)
printf(" ");
else
printf("\n");
}
}
for(int i = 0; i < Node[index].child.size(); i++){
int child = Node[index].child[i];
path[numNode] = child; //将child结点加到路径中
DFS(child, numNode + 1, sum + Node[child].weight);
}
}
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);
}
path[0] = 0; //路径的第一个结点设置为0号结点
DFS(0, 1, Node[0].weight);
return 0;
}