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当前节点的两种情况:
1.beChoosed = {son.beAbandoned乘积} //当前节点选中的情况下,子节点都不能选
2.beAbandoned = {(son.beAbandoned + son.beChoosed)乘积} //当前节点不选的情况下,子节点所有情况都算上
ps:最终要减去一种所有节点都未被选中的情况 -1
1 #include <iostream> 2 #include <stdio.h> 3 #include <list> 4 #define MOD 10007 5 #define MAXN 100002 6 using namespace std; 7 int n; 8 struct node{ 9 int parent; 10 list<int>son; 11 int son_count; 12 int beChoosedCount,beAbandonedCount; 13 }num[MAXN]; 14 15 void dfs(int id) 16 { 17 for(int i=0;i<num[id].son_count;++i) 18 { 19 int tmp = num[id].son.front(); 20 dfs(tmp); 21 num[id].son.pop_front(); 22 num[id].beChoosedCount *= num[tmp].beAbandonedCount%MOD; 23 num[id].beAbandonedCount *= (num[tmp].beChoosedCount+num[tmp].beAbandonedCount)%MOD; 24 } 25 26 } 27 int main() 28 { 29 cin>>n; 30 num[1].beChoosedCount=1; 31 num[1].beAbandonedCount=1; 32 for (int i = 2; i <= n; ++i) 33 { 34 35 cin>>num[i].parent; 36 num[num[i].parent].son.push_back(i); 37 num[num[i].parent].son_count++; 38 num[i].beAbandonedCount=1; 39 num[i].beChoosedCount=1; 40 } 41 42 dfs(1); 43 cout<<(num[1].beAbandonedCount+num[1].beChoosedCount-1)%MOD; 44 return 0; 45 }
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原文地址:http://www.cnblogs.com/sylvialucy/p/5032875.html