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

解析：用vector存樹，按照weight從大到小排序，再dfs中記錄路徑。

#include<set> #include<map> #include<list> #include<queue> #include<deque> #include<cmath> #include<stack> #include<cstdio> #include<string> #include<bitset> #include<vector> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #include<bits/stdc++.h> using namespace std;#define e exp(1) #define p acos(-1) #define mod 1000000007 #define inf 0x3f3f3f3f #define ll long long #define ull unsigned long long #define mem(a,b) memset(a,b,sizeof(a)) int gcd(int a,int b) {return b?gcd(b,a%b):a; }struct node{int weight;vector<int> children; }; vector<node> tree; vector<int> path; int n,m,w;bool cmp(int a,int b) {return tree[a].weight>tree[b].weight; }void dfs(int s,int sum) {if(sum>w)return ;if(sum==w){if(tree[s].children.size()!=0)return ;printf("%d",tree[0].weight);for(int i=0; i<path.size(); i++){printf(" %d",path[i]);}printf("\n");}for(int i=0; i<tree[s].children.size(); i++){int now=tree[s].children[i];path.push_back(tree[now].weight);dfs(now,sum+tree[now].weight);path.pop_back();} } int main() {scanf("%d%d%d",&n,&m,&w);tree.resize(n);for(int i=0; i<n; i++){scanf("%d",&tree[i].weight);}int id,k;for(int i=0; i<m; i++){scanf("%d%d",&id,&k);tree[id].children.resize(k);for(int j=0; j<k; j++){scanf("%d",&tree[id].children[j]);}sort(tree[id].children.begin(),tree[id].children.end(),cmp);}dfs(0,tree[0].weight);return 0; }

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