1053. Path of Equal Weight (30)

时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue

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 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 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 {A₁, A₂, ..., A_n} is said to be greater than sequence {B₁, B₂, ..., B_m} if there exists 1 <= k < min{n, m} such that A_i = B_ifor 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 纸老虎。挺简单的。只要把每个node里的孩子，按降序排一下就好。 #include<string> #include<stdio.h> #include<iostream> #include<string.h> #include<queue> #include<algorithm> #include<map> #include<vector> using namespace std; int value[101]; struct { int cnt=0; int child[100]; }node[101]; void paixu(int x) { int cnt=node[x].cnt; if(cnt==0) return; for(int i=1;i<cnt;i++) for(int j=0;j<cnt-i;j++) if(value[node[x].child[j]]<value[node[x].child[j+1]]) swap(node[x].child[j],node[x].child[j+1]); } int path[101]; int pp=0; long long sum=0; int n,m,s; void dfs(int x) { sum+=value[x]; path[pp++]=value[x]; if(sum==s) { if(node[x].cnt==0) { int flag=0; for(int i=0;i<pp;i++) { if(flag==1) cout<<' '; flag=1; cout<<path[i]; } cout<<endl; } pp--; sum-=value[x]; return; } if(sum>s) { pp--; sum-=value[x]; return; } for(int i=0;i<node[x].cnt;i++) dfs(node[x].child[i]); pp--; sum-=value[x]; } int main() { cin>>n>>m>>s; for(int i=0;i<n;i++) cin>>value[i]; for(int i=0;i<m;i++) { int root,cnt; cin>>root>>cnt; node[root].cnt=cnt; for(int j=0;j<cnt;j++) cin>>node[root].child[j]; } for(int i=0;i<n;i++) paixu(i); dfs(0); return 0; }