1053 Path of Equal Weight
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, the number of nodes in a tree, M (<), the number of non-leaf nodes, and 0, the given weight number. The next line contains N positive numbers where Wi (<) 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 { is said to be greater than sequence { if there exists 1 such that Ai=Bi for ,, 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
#include<iostream> #include<string> #include<algorithm> using namespace std; const int maxn=100; struct node{ int id; int value; int num1=0; int sub[maxn]; }; node str[maxn]; int a[maxn]; int n,m,w,flag=0; string s[maxn]; int cmp(string s1,string s2) { return s1>s2; } void dfs(int id1,int w1,int cnt) { if(w1<0) return ; if(w1==0&&str[id1].num1==0) { s[flag]=""; for(int i=0;i<cnt;i++) { s[flag]=s[flag]+char(a[i]+'0'); } //cout<<flag<<endl; //cout<<s[flag]<<endl; flag++; } for(int i=0;i<str[id1].num1;i++) { a[cnt]=str[str[id1].sub[i]].value; dfs(str[id1].sub[i],w1-str[str[id1].sub[i]].value,cnt+1); } } int main() { int k,k1,temp; cin>>n>>m>>w; for(int i=0;i<n;i++) { cin>>str[i].value; } while(m--) { cin>>k>>k1; str[k].num1=k1; for(int i=0;i<k1;i++) { cin>>temp; str[k].sub[i]=temp; } } a[0]=str[0].value; dfs(0,w-str[0].value,1); sort(s,s+flag,cmp); //cout<<flag<<endl; for(int i=0;i<flag;i++) { for(int j=0;j<s[i].size();j++) { if(j==0) cout<<int(s[i][j]-'0'); else cout<<" "<<int(s[i][j]-'0'); } cout<<endl; } return 0; }
总有一些case过不了