[codeforces-543-D div1]树型DP
题意:给一棵树的边标上0或1,求以节点i为源点,其它点到i的唯一路径上的1的边数不超过1条的方案数,输出所有i的答案。
思路:令f[i]表示以节点i为源点,只考虑子树i时的方案数,ans[i]为最后答案,fa[i]为i的父亲,则不难得出以下转移方程:
f[i] = ∏(1 + f[v]),v是i的儿子 ans[i] = f[i] * (1 + ans[fa[i]] / (1 + f[i]))
由于除法取模运算的存在,不得不对1+f[i]求逆元,但1+f[i]可能等于MOD,于是这种情况下结果就是错的了,不能用这个公式求。
令g[i] = ans[fa[i]] / (1 + f[i]),注意到g[i]实际上等于∏(1 + f[v]) * g[fa[i]],v是i的兄弟,于是可以增加一个前缀积数组和一个后缀积数组用来得到∏(1 + f[v]),就不难得到g[i]和最后的答案了。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define define_m int m = (l + r) >> 1 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 32 #define rep_down1(a, b) for (int a = b; a > 0; a--) 33 #define all(a) (a).begin(), (a).end() 34 #define lowbit(x) ((x) & (-(x))) 35 #define constructInt5(name, a, b, c, d, e) name(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0): a(a), b(b), c(c), d(d), e(e) {} 36 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 37 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 38 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 39 #define pchr(a) putchar(a) 40 #define pstr(a) printf("%s", a) 41 #define sstr(a) scanf("%s", a) 42 #define sint(a) scanf("%d", &a) 43 #define sint2(a, b) scanf("%d%d", &a, &b) 44 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 45 #define pint(a) printf("%d\n", a) 46 #define test_print1(a) cout << "var1 = " << a << endl 47 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 48 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 49 #define mp(a, b) make_pair(a, b) 50 #define pb(a) push_back(a) 51 52 typedef long long LL; 53 typedef pair<int, int> pii; 54 typedef vector<int> vi; 55 56 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 57 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 58 const int maxn = 2e5 + 7; 59 const int md = 1e9 + 7; 60 const int inf = 1e9 + 7; 61 const LL inf_L = 1e18 + 7; 62 const double pi = acos(-1.0); 63 const double eps = 1e-6; 64 65 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 66 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 67 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 68 template<class T>T condition(bool f, T a, T b){return f?a:b;} 69 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 70 int make_id(int x, int y, int n) { return x * n + y; } 71 72 struct Graph { 73 vector<vector<int> > G; 74 void clear() { G.clear(); } 75 void resize(int n) { G.resize(n + 2); } 76 void add(int u, int v) { G[u].push_back(v); } 77 vector<int> & operator [] (int u) { return G[u]; } 78 }; 79 Graph G, pre, suf; 80 int fa[maxn], f[maxn], ans[maxn], id[maxn], g[maxn]; 81 82 void dfs(int n) { 83 f[n] = 1; 84 int sz = G[n].size(); 85 rep_up0(i, sz) { 86 int v = G[n][i]; 87 dfs(v); 88 f[n] = (LL)f[n] * (1 + f[v]) % md; 89 } 90 } 91 92 void getAns(int n) { 93 int sz = G[n].size(); 94 int fn = fa[n], in = id[n]; 95 pre[n].resize(sz + 2); 96 suf[n].resize(sz + 2); 97 pre[n][0] = suf[n][sz + 1] = 1; 98 if (n == 1) { 99 ans[n] = f[n]; 100 g[n] = 1; 101 } 102 else { 103 g[n] = (1 + (LL)pre[fn][in - 1] % md * suf[fn][in + 1] % md * g[fn]) % md; 104 ans[n] = (LL)f[n] * g[n] % md; 105 } 106 rep_up0(i, sz) { 107 int v = G[n][i]; 108 pre[n][i + 1] = (LL)pre[n][i] * (1 + f[v]) % md; 109 } 110 rep_down0(i, sz) { 111 int v = G[n][i]; 112 suf[n][i + 1] = (LL)suf[n][i + 2] * (1 + f[v])% md; 113 } 114 rep_up0(i, sz) { 115 int v = G[n][i]; 116 getAns(v); 117 } 118 } 119 120 int main() { 121 //freopen("in.txt", "r", stdin); 122 int n; 123 cin >> n; 124 G.resize(n); 125 pre.resize(n); 126 suf.resize(n); 127 for (int i = 2; i <= n; i ++) { 128 int x; 129 sint(x); 130 G.add(x, i); 131 fa[i] = x; 132 id[i] = G[x].size(); 133 } 134 dfs(1); 135 getAns(1); 136 rep_up1(i, n) printf("%d%c", ans[i], i == n? '\n' : ' '); 137 return 0; 138 }