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; }