luogu P2016 战略游戏

嘟嘟嘟

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

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = (ans << ) + (ans << ) + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n;

 struct Edge
 {
     int nxt, to;
 }e[maxn << ];
 int head[maxn], ecnt = -;
 void addEdge(int x, int y)
 {
     e[++ecnt] = (Edge){head[x], y};
     head[x] = ecnt;
 }

 ll dp[maxn][];
 void dfs(int now, int f)
 {
     dp[now][] = ;
     for(int i = head[now], v; i != -; i = e[i].nxt)
     {
         v = e[i].to;
         if(v == f) continue;
         dfs(v, now);
         dp[now][] += dp[v][];
         dp[now][] += min(dp[v][], dp[v][]);
     }
 }

 int main()
 {
     Mem(head, -);
     n = read();
     for(int i = ; i <= n; ++i)
     {
         int x = read(), k = read(); x++;
         for(int j = ; j <= k; ++j)
         {
             int y = read(); y++;
             addEdge(x, y); addEdge(y, x);
         }
     }
     dfs(, );
     write(min(dp[][], dp[][])), enter;
     return ;
 }