1053 Path of Equal Weight (30 分)(樹的遍歷)
阿新 • • 發佈:2018-12-07
給定一棵樹和每個結點的權值,求所有從根結點到葉子結點的路徑,讓每條路徑上的結點的權值之和等於給定的常數,如果有多條這樣的路徑,按照非遞增的順序輸出
#include<iostream> #include<vector> #include<algorithm> using namespace std; int n, m, s; const int maxn = 110; struct node { int weight; vector<int>child; }Node[maxn]; vector<int>ans;bool comp(int a, int b) { return Node[a].weight > Node[b].weight; } void DFS(int index, int sum) { if (sum == s) { if (Node[index].child.size() != 0)return; for (int i = 0; i < ans.size(); i++) { printf("%d", Node[ans[i]].weight); if (i != ans.size() - 1)printf(" "); } printf("\n"); return; } if (sum > s)return; for (int i = 0; i < Node[index].child.size(); i++) { int child = Node[index].child[i]; ans.push_back(child); DFS(child, sum + Node[child].weight); ans.pop_back(); } }int main() { scanf("%d%d%d", &n, &m, &s); for (int i = 0; i < n; i++)scanf("%d", &Node[i].weight); while (m--) { int id, k; scanf("%d%d", &id, &k); while (k--) { int idk; scanf("%d", &idk); Node[id].child.push_back(idk); } sort(Node[id].child.begin(), Node[id].child.end(), comp); } ans.push_back(0); DFS(0, Node[0].weight); return 0; }