Problem

给你一棵树,可以在每个点上选择造塔或不造,每座塔可以覆盖这个节点和相邻节点,问覆盖整棵树的最小塔数。

Solution

看到这道题的第一眼,我就觉得是一题贪心题,但看见出题的时候分类在树形DP,于是就没仔细想贪心。
树形DP:f[u][0]表示u被其儿子覆盖,f[u][1]表示u上有塔,f[u][2]表示u被其父亲覆盖,转移显然
贪心:我们dfs到叶子节点时,尽量贪心覆盖它的父亲

Notice

树形DP的状态确实很难想到

Code

树形DP

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; ++i)
#define per(i, a, b) for (reg i = a; i >= b; --i)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 10000;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
struct node
{
    int vet, next;
}edge[2 * N + 5];
int ans = 0, flag[N + 5], num = 0, head[N + 5];
inline void add(const int &u, const int &v)
{
    edge[++num].vet = v, edge[num].next = head[u], head[u] = num;
    edge[++num].vet = u, edge[num].next = head[v], head[v] = num;
}
inline void dfs(reg u, reg fa)
{
    reg tt = 0;
    travel(i, u)
    {
        int v = edge[i].vet;
        if (v == fa) continue;
        dfs(v, u);
        if (flag[v]) tt = 1;
    }
    if (!tt && !flag[u] && !flag[fa])
    {
        ans++;
        flag[fa] = 1;
    }
}
int sqz()
{
    reg n = read();
    rep(i, 1, n - 1) add(read(), read());
    dfs(1, 0);
    write(ans); puts("");
    return 0;
}

贪心

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e8, N = 10000;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
int f[N + 5][3], head[N + 5];
int num = 0;
struct node
{
	int vet, next;
}edge[2 * N + 5];
void add(int u, int v)
{
	edge[++num].vet = v;
	edge[num].next = head[u];
	head[u] = num;
	edge[++num].vet = u;
	edge[num].next = head[v];
	head[v] = num;
}
void dp(int u, int fa)
{
	int sum = 0;
	f[u][1] = 1, f[u][0] = INF;
	travel(i, u)
	{
		int v = edge[i].vet;
		if (v == fa) continue;
		dp(v, u);
		f[u][0] = min(f[u][0], f[v][1] - min(f[v][1], f[v][0]));
		f[u][1] += min(f[v][0], min(f[v][1], f[v][2]));
		f[u][2] += f[v][0];
		sum += min(f[v][0], f[v][1]);
	}
	f[u][0] += sum;
}
int sqz()
{
	int n = read();
	rep(i, 1, n - 1) add(read(), read());
	dp(1, 0);
	printf("%d\n", min(f[1][0], f[1][1]));
	return 0;
}
posted on 2017-11-04 22:04  WizardCowboy  阅读(117)  评论(0编辑  收藏  举报