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-8;
const int maxn = 1505;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int n;

struct Edge
{
    int nxt, to;
}e[maxn << 1];
int head[maxn], ecnt = -1;
void addEdge(int x, int y)
{
    e[++ecnt] = (Edge){head[x], y};
    head[x] = ecnt;
}

ll dp[maxn][2];
void dfs(int now, int f)
{
    dp[now][1] = 1;
    for(int i = head[now], v; i != -1; i = e[i].nxt)
    {
        v = e[i].to;
        if(v == f) continue;
        dfs(v, now);
        dp[now][0] += dp[v][1];
        dp[now][1] += min(dp[v][0], dp[v][1]);
    }
}

int main()
{
    Mem(head, -1);
    n = read();
    for(int i = 1; i <= n; ++i)
    {
        int x = read(), k = read(); x++;
        for(int j = 1; j <= k; ++j)
        {
            int y = read(); y++;
            addEdge(x, y); addEdge(y, x);
        }
    }
    dfs(1, 0);
    write(min(dp[1][0], dp[1][1])), enter;
    return 0;
}