ZOJ 3824Fiber-optic Network 树形dp 2014牡丹江现场赛F

Fiber-optic Network

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Time Limit: 15 Seconds      Memory Limit: 262144 KB

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Marjar University has decided to upgrade the infrastructure of school intranet by using fiber-optic technology. There are N buildings in the school. Each building will be installed with one router. These routers are connected by optical cables in such a way that there is exactly one path between any two routers.

Each router should be initialized with an operating frequency F_i before it starts to work. Due to the limitations of hardware and environment, the operating frequency should be an integer number within [L_i, R_i]. In order to reduce the signal noise, the operating frequency of any two adjacent routers should be co-prime.

Edward is the headmaster of Marjar University. He is very interested in the number of different ways to initialize the operating frequency. Please write a program to help him! To make the report simple and neat, you only need to calculate the sum of F_i (modulo 1000000007) in all solutions for each router.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains one integer N (1 <= N <= 50). The next line contains N integers L_i (1 <= L_i <= 50000). Then, the following line contains N integers R_i (L_i <= R_i<= 50000).

For the next N - 1 lines, each line contains two integers X_i and Y_i. That means there is an optical cable connecting router X_i and router Y_i (indexes are 1-based).

Output

For each test case, output a line with N integers representing the sum of F_i (modulo 1000000007) in all solutions.

Sample Input

241 2 3 42 3 4 51 22 33 441 2 3 42 3 4 51 21 31 4

Sample Output

5 10 14 1910 23 31 41

Hint

In the first sample test case, there are 4 ways to initialize the operating frequency:

• 1 2 3 4
• 1 2 3 5
• 1 3 4 5
• 2 3 4 5

题意：有一颗树，要为每个节点赋一个值Fi，使得任意相邻的节点互素。然后对每个节点统计Fi在所有可能中的和。

思路：其实题目要求的就是对于每个点我们赋值为X的时候有多少种方案，然后每个点的答案就是 sigma(x * (对应方案数))。好了，我们的注意力转移到如何求方案数。

我们先看下当树是一条链的时候的情况，因为这种情况比较简单。 我们可以容易得到转移方程dp[i][j] = sigma(dp[i-1][k]) (k 与 j互素) 其中dp[i][j]表示前i个节点，第i个节点赋为j时的方案数。 但是显然复杂度已经是 50 * 50000 * 50000 ，太大了，不能忍！！而且我们能注意到其实互素的个数是很多的，我们不妨先求出跟j不是互素的，然后用总的减掉不是互素的方案数。如果是求不是互素要怎么求呢？我们能够用容斥，我的代码里貌似是类似莫比乌斯反演的东西？ 我们把j进行分解素因子，然后枚举一下因子的组合容斥，选奇数个因子就是加上，偶数个就是减去（如果不懂我说什么可以先了解一下容斥的思想，并不是简单的全集减去对立面）。

如果要这么做，我们的状态的意义就要改一下了，这时候dp[i][j]中的j表示的是第i个节点的数字能被j整除时的方案数，我们继续注意到对每个赋的值x进行容斥统计的时候，我们只关注它有什么素因子，所以我们能继续把一些素因子完全一样的数一起进行统计。于是我们在节点i枚举每个素因子组成不同的数x来统计方案数S，然后把S加到每个dp[i][k] （k是x因子，因为我们的状态表示的是该位置的数字能被k整除时的方案数）。这样单链的情况我们基本就做出来了，当是树的时候一样的，只不过方案数是各个分支的方案数的乘积而已。

好了，现在如果知识要求出一个点Fi的和很简单是吧？但是题目要求所有的点的Fi和。我们当然可以直接以每个点作为根来做一遍dp，我这样做之间简单粗暴的返回tle。我们能不能把那个n*n的复杂度优化为n呢？是可以的。我们考虑一个题目：在一颗树里，求出每个点的最远节点的距离 。我们可以用dp[u][fa]表示在把(u,fa)这条边砍掉后，以u为根的子树中的“最远距离”，那么我们的递推就是dp[u][fa] = max(dp[v][u] + 1) (v是u的儿子）。 这里我们能够用记忆化搜索来做。 回到这道题，可以用一样的处理方法。不过直接多加一维会爆内存的，其实那个二维的状态我们可以用O(n)的空间存的，因为每一条边对应两个状态，所以状态数只有2*n-2. 我们另外开一个id[maxn][maxn]对状态进行标记就行了。具体的可以看代码。

代码：

#include <iostream>#include <vector>#include <algorithm>#include <string.h>#include <cstring>#include <stdio.h>#include <cmath>#include <math.h>#define rep(i,a,b) for(int i=(a);i<(b);++i)#define rrep(i,b,a) for(int i = (b); i >= (a); --i)#define clr(a,x) memset(a,(x),sizeof(a))#define LL long longusing namespace std;const int maxn = 50 + 5;const int maxv = 50000 + 5;const int mod = 1e9 + 7;LL num[maxn * 2][maxv];LL Count[maxn * maxn];int id[maxn][maxn];vector<int> pe[maxv];vector<int> fac[maxv];  int same[maxn][maxv];int h[maxv];int miu[maxv];int n;int l[maxn], r[maxn];vector<int> kind[maxn];vector<int> adj[maxn];void add(LL & a, LL b){a += b;if (a >= mod) a -= mod;}LL Cal_num(LL * arr, int x, LL amt){rep(i, 0, fac[x].size()) {int z = fac[x][i];amt += arr[z] * miu[z];if (amt >= mod) amt -= mod;if (amt < 0) amt += mod;}return amt;}bool vis[maxn][maxn];void dfs(int u, int fa){    if (vis[u][fa]) return;bool isleaf = true;vis[u][fa] = true;rep(i, 0, adj[u].size()) {int v = adj[u][i];if (v == fa) continue;isleaf = false;dfs(v, u);}if (isleaf) {rep(d, 0, kind[u].size()) {    int i = kind[u][d];rep(j, 0, fac[i].size()) {int x = fac[i][j];num[id[u][fa]][x] += same[u][i];}Count[id[u][fa]] += same[u][i];}}else {//统计方案数LL way;rep(e, 0,kind[u].size()) {int d = kind[u][e];way = same[u][d];rep(i, 0, adj[u].size()) {int v = adj[u][i];if (v == fa) continue;way = (way * Cal_num(num[id[v][u]], d, Count[id[v][u]])) % mod;}rep(j, 0, fac[d].size()) {int x = fac[d][j];add(num[id[u][fa]][x], way);}add(Count[id[u][fa]], way);}}}void solve(){    clr(id,-1);    clr(Count,0); clr(num,0); clr(vis,0);    int c = 0;    rep(i,1,n+1) {        rep(j,0,adj[i].size()) {            id[i][adj[i][j]] = ++c;        }//        id[i][i] = ++c;    }    rep(u,1,n+1) {        rep(j,0,adj[u].size()) {            int v = adj[u][j];            dfs(u,v);        }//        dfs(u,u);    }rep(u, 1, n + 1) {LL ans = 0;rep(i, l[u], r[u] + 1) {LL way = 1;rep(j, 0, adj[u].size()) {int v = adj[u][j];way = (way * Cal_num(num[id[v][u]], i, Count[id[v][u]])) % mod;}add(ans, way * i % mod);}if (u != 1) printf(" ");printf("%lld", ans);}puts("");}void read_int(int & x){    char ch = getchar();    while (ch < '0' || ch > '9') ch = getchar();    x = ch - '0'; ch = getchar();    while ('0' <= ch && ch <= '9') {        x = 10 * x + ch - '0';        ch = getchar();    }}void Getinput(){freopen("in.txt", "w", stdout);int T = 1; printf("%d\n", T);while (T--) {int n = 50; printf("%d\n", n);rep(i, 0, n) printf("1 ");puts("");rep(i, 0, n) printf("50000 ");puts("");rep(i, 0, n - 1) printf("%d %d\n", i + 1, i + 2);}}int main(){//Getinput(); return 0;//freopen("in.txt", "r", stdin);//    freopen("out.txt","w",stdout);rep(i, 1, maxv) {int x = i;if (~x & 1) {pe[i].push_back(2);while (~x & 1) x >>= 1;}for (LL j = 3; j*j <= x; j += 2) if (x % j == 0) {pe[i].push_back(j);do x /= j;while (x % j == 0);}if (x > 1) pe[i].push_back(x);//printf("%u\n", fac[i].size());int k = pe[i].size();int maxx = 1;rep(s, 1, 1 << k) {int x = 1;int odd = 0;rep(j, 0, k) if (s & (1 << j)) x *= pe[i][j], odd ^= 1;if (x > maxx) maxx = x;miu[x] = odd ? -1 : 1;fac[i].push_back(x);}h[i] = maxx;//printf("%u\n", fac[i].size());}int T; cin >> T;while (T--) {        read_int(n);//scanf("%d", &n);rep(i, 1, n + 1) read_int(l[i]); //scanf("%d", l + i);rep(i, 1, n + 1) read_int(r[i]); //scanf("%d", r + i);clr(same, 0);rep(i, 1, n + 1) {kind[i].clear();rep(j, l[i], r[i] + 1) {if (++same[i][h[j]] == 1) kind[i].push_back(h[j]);}//printf("%u\n",kind[i].size());}rep(i, 0, maxn) adj[i].clear();rep(i, 1, n) {int u, v; read_int(u); read_int(v); //scanf("%d%d", &u, &v);adj[u].push_back(v);adj[v].push_back(u);}solve();}}