Cell Phone Networ (树形dp-最小支配集)

梦里梦外; 2023-06-01 06:12 21阅读 0赞

*  Cell Phone Networ (树形dp-最小支配集)
    
     *  题意
     *  思路
     *  题解

# Cell Phone Networ (树形dp-最小支配集) #

Farmer John has decided to give each of his cows a cell phone in hopes to encourage their social interaction. This, however, requires him to set up cell phone towers on his N (1 ≤ N ≤ 10,000) pastures (conveniently numbered 1..N) so they can all communicate.

Exactly N-1 pairs of pastures are adjacent, and for any two pastures A and B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B) there is a sequence of adjacent pastures such that A is the first pasture in the sequence and B is the last. Farmer John can only place cell phone towers in the pastures, and each tower has enough range to provide service to the pasture it is on and all pastures adjacent to the pasture with the cell tower.

Help him determine the minimum number of towers he must install to provide cell phone service to each pasture.

**Input**

*  Line 1: A single integer: N
 *  Lines 2..N: Each line specifies a pair of adjacent pastures with two space-separated integers: A and B

**Output**

*  Line 1: A single integer indicating the minimum number of towers to install

**Sample Input**  
5  
1 3  
5 2  
4 3  
3 5  
**Sample Output**  
2

## 题意 ##

任意一点都要有信号塔或者在信号塔周围（信号塔范围为 *1* ）。那么至少要有多少信号塔。

## 思路 ##

首先确定几种状态

*  **dp\[x\]\[0\]** :不放塔但是**在范围内**
 *  **dp\[x\]\[1\]** :不放塔且**不在范围内**
 *  **dp\[x\]\[2\]** :放塔

由于是由子节点推父节点，故下面在计算某点状态的时候，只考虑其子节点的状态。因为，只需计算出当前节点自己的所有状态，而用哪种状态是父节点的事。

1.  对于不放塔但是在范围内的，肯定是某个子节点有塔。  
    但是所有子节点为了最优，都没有选择塔，那怎么办。  
    这时候就需要，选择一个**变成有塔的差值最小的一个子节点**。所以在遍历子节点时，就要记录下**放塔与不放塔**最小差值，和**changed**标记。如果需要，就加上差值。
2.  对于不放塔但是不在范围内的，只要其继续错下去就好了。但是，这是只考虑子节点的情况。在父节点放塔的时候，这个点可能又变成在信号内了，所以这种状态要保证其所有子节点都是合理的。这样父节点只要考虑这个节点就好，不需要关心下面的节点了。  
    这样得**dp\[now\]\[1\]+=dp\[son\]\[0\]** 为什么是dp\[son\]\[**0**\]，因为子节点合理并且子节点没有信号塔啊。
3.  对于放塔的，子节点怎么放都行，所以得**dp\[u\]\[2\] += min(dp\[v\]\[0\], min(dp\[v\]\[1\], dp\[v\]\[2\]))**
4.  当然别忘了初始化每个节点。

## 题解 ##

#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <algorithm> #define N 100005 #define INF 0x3f3f3f using namespace std; vector<int> eg[N]; int dp[N][3], foo[N], vis[N]; int n, x, y; void dfs(int u) { dp[u][0] = dp[u][1] = 0; dp[u][2] = 1; vis[u] = 1; int minmall = INF, f = 0; for (int i = 0; i < eg[u].size(); i++) { int v = eg[u][i]; if (vis[v]) continue; dfs(v); minmall = (minmall, dp[v][2] - dp[v][0]); //记录差值 if (dp[v][2] <= dp[v][0]) f = 1; //子节点能放就放，应为还能照顾父节点。所以这里是 <= //如果子节点有放塔的，就标记，差值就无需在加了。 dp[u][0] += min(dp[v][2], dp[v][0]); dp[u][1] += dp[v][0]; dp[u][2] += min(dp[v][0], min(dp[v][1], dp[v][2])); } if (f == 0)//如果没有子节点放塔，加上使差值最小的节点放上塔。 dp[u][0] += minmall; } int main() { while (cin >> n) { memset(dp, 0, sizeof dp); memset(vis, 0, sizeof vis); memset(foo, -1, sizeof foo); for (int i = 0; i < n - 1; i++) { cin >> x >> y; eg[x].push_back(y); eg[y].push_back(x); foo[y] = x; } int root = 1; while (foo[root] != -1) root = foo[root]; vis[root] = 1; dfs(root); cout << min(dp[root][0], dp[root][2]) << endl; for (int i = 1; i <= n; i++) eg[i].clear(); } }

转载于:https://www.cnblogs.com/tttfu/p/11293850.html