1053 Path of Equal Weight (30分)
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<2
30
, 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 {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
思路還是很簡單的，深度搜索搞定，不過需要注意邊如何保存，以及如何判斷葉子節點，結果如何保存，結果如何排序
使用vector<vector >vec(101)來存儲邊，權重使用int weight[101]存儲。搜索時候使用一個vector out, 先把根節點0存進去，然後權重值減去根節點的權重，調用df算法。葉子結點判斷可以使用vec[節點編號].size()==0
排序可以使用自己寫一個sort排序
#include <iostream>
#include <algorithm>
using namespace std;
#include <vector>
vector<vector<int>> edge(101);
int weight[101];
vector<int> out;
vector<vector<int>> beforSort;
void df(int root,int lastSum)
{
if(lastSum==0&&edge[root].size()==0)
{
beforSort.push_back(out);
return ;
}
else if(lastSum<0)
{
return;
}
else
{
for(int i=0;i<edge[root].size();i++){
out.push_back(edge[root][i]);
df(edge[root][i],lastSum-weight[edge[root][i]]);
out.pop_back();
}
}
}
bool sort2(vector<int> a,vector<int> b ){
int i=0;
while(i<a.size()&&i<b.size()){
int aa=weight[a[i]],bb=weight[b[i]];
if(aa>bb){
return 1;
}
else if(aa<bb)
return 0;
i++;
}
return 0;
}
int main()
{
int N,inNum,SUM;
cin>>N>>inNum>>SUM;
for(int i=0; i<N; i++)
{
cin>>weight[i];
}
for(int i=0; i<inNum; i++)
{
int parent;
cin>>parent;
int k;
cin>>k;
for(int p=0; p<k; p++)
{
int son;
cin>>son;
edge[parent].push_back(son);
}
}
//0是根節點
out.push_back(0);
df(0,SUM-weight[0]);
sort(beforSort.begin(),beforSort.end(),sort2);
//輸出vec元素對應的weight
for(int q=0;q<beforSort.size();q++){
for(int i=0;i<beforSort[q].size();i++){
if(i==0){
cout<<weight[beforSort[q][i]];
}
else{
cout<<' '<<weight[beforSort[q][i]];
}
}
cout<<endl;
}
return 0;
}