题目
题目描述
Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?
Your program should find the minimum number of soldiers that Bob has to put for a given tree.
For example for the tree:
the solution is one soldier ( at the node 1).
输入描述:
The input contains several data sets in text format. Each data set represents a tree with the following description:
the number of nodes
the description of each node in the following format
node_identifier:(number_of_roads) \(node\_identifier_1\) \(node\_identifier_2\) ... \(node\_identifier_{number\_of\_roads }\)
or
node_identifier:(0)
The node identifiers are integer numbers between 0 and n-1, for n nodes \((0 \lt n \leq 1500)\) ;the number_of_roads in each line of input will no more than 10. Every edge appears only once in the input data.
输出描述
The output should be printed on the standard output. For each given input data set, print one integer number in a single line that gives the result (the minimum number of soldiers). An example is given in the following:
示例1
输入
4
0:(1) 1
1:(2) 2 3
2:(0)
3:(0)
5
3:(3) 1 4 2
1:(1) 0
2:(0)
0:(0)
4:(0)
输出
1
2
题解
知识点：树形dp
题目要求最少点覆盖所有边（最小点覆盖），一个点能覆盖所连的所有边，所以有如下情况。
以 \(1\) 为根，设 \(dp[u][0/1]\) 表示以 \(u\) 为根的子树，\(u\) 的状态是不选/选的最小值。转移方程为：
\[\left \{ \begin{array}{l} dp[u][0] = \sum dp[v_i][1]\\ dp[u][1] = \sum \min(dp[v_i][0],dp[v_i][1]) \end{array} \right . \]表示 \(u\) 不选则孩子必须选；\(u\) 选了孩子可选可不选，取最小值。
时间复杂度 \(O(n)\)
空间复杂度 \(O(n)\)
代码
#include <bits/stdc++.h>
using namespace std;
vector<int> g[1507];
int dp[1507][2];
///快读
template<class T>
inline void read(T &val) {
T x = 0, f = 1;char c = getchar();
while (c < '0' || c>'9') { if (c == '-') f = -1;c = getchar(); }///整数符号
while (c >= '0' && c <= '9') { x = (x << 3) + (x << 1) + (c ^ 48);c = getchar(); }///挪位加数
val = x * f;
}
void dfs(int u, int fa) {
for (auto v : g[u]) {
if (v == fa) continue;
dfs(v, u);
dp[u][0] += dp[v][1];
dp[u][1] += min(dp[v][0], dp[v][1]);
}
dp[u][1]++;
}
int main() {
std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n;
while (~scanf("%d", &n)) {
memset(dp, 0, sizeof(dp));
for (int u = 0;u < n;u++) g[u].clear();
for (int i = 1;i <= n;i++) {
int u, cnt;
read(u);
read(cnt);
for (int j = 1, v;j <= cnt;j++) {
read(v);
g[u].push_back(v);
g[v].push_back(u);
}
}
dfs(0, -1);
cout << min(dp[0][0], dp[0][1]) << '\n';
}
return 0;
}