树形dp水题啦。
刚开始以为和[SDOI2006]保安站岗这道题一样，然后交上去WA了。
仔细想想还是有区别的，一个是能看到相邻点，一个是能看到相邻边。对于第一个，可以(u, v)两个点都不放，然而对于这道题就不行了。
不过dp方程更简单：dp[u][0/1]表示u这个点不放/放士兵的最优答案。那么：
不放：则u的所有儿子必须放，才能覆盖所有相邻的边：dp[u][0] = Σdp[v][1]
放：则u的儿子可放可不放：dp[u][1] = Σmin{dp[v][0], dp[v][1]} + 1
完啦
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 1505; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n; 38 39 struct Edge 40 { 41 int nxt, to; 42 }e[maxn << 1]; 43 int head[maxn], ecnt = -1; 44 void addEdge(int x, int y) 45 { 46 e[++ecnt] = (Edge){head[x], y}; 47 head[x] = ecnt; 48 } 49 50 ll dp[maxn][2]; 51 void dfs(int now, int f) 52 { 53 dp[now][1] = 1; 54 for(int i = head[now], v; i != -1; i = e[i].nxt) 55 { 56 v = e[i].to; 57 if(v == f) continue; 58 dfs(v, now); 59 dp[now][0] += dp[v][1]; 60 dp[now][1] += min(dp[v][0], dp[v][1]); 61 } 62 } 63 64 int main() 65 { 66 Mem(head, -1); 67 n = read(); 68 for(int i = 1; i <= n; ++i) 69 { 70 int x = read(), k = read(); x++; 71 for(int j = 1; j <= k; ++j) 72 { 73 int y = read(); y++; 74 addEdge(x, y); addEdge(y, x); 75 } 76 } 77 dfs(1, 0); 78 write(min(dp[1][0], dp[1][1])), enter; 79 return 0; 80 }
