A1053 Path of Equal Weight (30 point(s))

DFS 取和为s的路径

1. 原文

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 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 {$A_1$,$A_2$,⋯,$A_n$} is said to be greater than sequence {$B_1$$B_2$,⋯,$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

2. 解析

按权重从高到低排序，return v[a].weight>v[b].weight;

和为s if (sum==s)

记录路径 path[cnt]=idx;

dfs(temp, sum+v[temp].weight,index+1);

3. AC代码

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
struct node
{
	int weight;
	vector<int> children;
}v[110];
bool cmp(int a,int b){
	return v[a].weight>v[b].weight;
}
int s;
int path[110]={};
void dfs(int root,int sum,int cnt){
	//printf("%d %d\n",root,sum);
	if (sum>s)
	{
		return;
	}
	if (sum==s)
	{
		if (v[root].children.size()!=0)
		{
			return;
		}
		for (int i = 0; i < cnt; ++i)
		{
			if (i!=0)
			{
				printf(" ");
			}
			printf("%d",v[path[i]].weight);
		}
		printf("\n");
		return;
	}
	for (int i = 0; i < v[root].children.size(); ++i)
	{
		int idx=v[root].children[i];
    path[cnt]=idx;
		dfs(idx,sum+v[idx].weight,cnt+1);
	}

}
int main()
{
	int n,m;
	scanf("%d%d%d",&n,&m,&s);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&v[i].weight);
	}
	int id,k,a;
	for (int i = 0; i < m; ++i)
	{
		scanf("%d%d",&id,&k);
		for (int j = 0; j < k; ++j)
		{
			scanf("%d",&a);
			v[id].children.push_back(a);
		}
		sort(v[id].children.begin(),v[id].children.end(),cmp);
	}
	path[0]=0;
	dfs(0,v[0].weight,1);
	return 0;
}