Path of Equal Weight (DFS)
Path of Equal Weight (DFS)
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 Figure 1: 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 Figure 1.
Figure 1
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
这道30分的题目，提交一次就意外的AC了。
就是 建立连接表 DFS+记录路径+权值累加 搜到叶子节点，如果权值之和与要求的的相等时保存路径。
最后的排序要点混，但进行三层的判断排序，也就能过了，
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int WW[100];
int visit[100];
vector<int> vv[100];
vector<int> road;
vector<int> RR[100];
int sum,wi;
bool cmp(vector<int> a,vector<int> b)
{
if(a[0]==b[0]&&a[1]==b[1])
return a[2]>b[2];
if(a[0]==b[0])
return a[1]>b[1];
return a[0]>b[0];
}
void DFS(int root,int &count)
{
if(visit[root]==0)
{
visit[root]=1;
road.push_back(WW[root]);
sum+=WW[root];
for(int i=0;i<vv[root].size();i++)
{
if(visit[vv[root][i]]==0)
DFS(vv[root][i],count);
}
if(sum==wi&&vv[root].size()==0)
{
RR[count++]=road;
}
road.pop_back();
sum-=WW[root];
}
}
int main()
{
int i,j,num,fnum;
while(cin>>num)
{
road.clear();
cin>>fnum>>wi;
for(i=0;i<num;i++)
{
cin>>WW[i];
vv[i].clear();
visit[i]=0;
RR[i].clear();
}
for(i=0;i<fnum;i++)
{
int n1,n2;
cin>>n1>>n2;
for(j=0;j<n2;j++)
{
int tem;
cin>>tem;
vv[n1].push_back(tem);
}
}
int count=0;
sum=0;
DFS(0,count);
sort(RR,RR+count,cmp);
for(i=0;i<count;i++)
{
cout<<RR[i][0];
for(j=1;j<RR[i].size();j++)
cout<<" "<<RR[i][j];
cout<<endl;
}
}
return 0;
}