2016长乐夏令营 Day13

T1：

如果x >= y 那么答案显然(n-1)*y 不过要特判菊花图

如果x < y，问题就变成在一棵树上尽可能多的选择链

f[i][k]代表以i为根的子树，i连边的状态为k，选的最多的边的数量

k <= 2

转移显然

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

typedef long long LL;
const int maxn = 1E6 + 10;
const int INF = 1E9;

int n,du[maxn];
LL x,y,ans,f[maxn][3];

vector  v[maxn];

int dfs(int x,int fa)
{
	f[x][0] = f[x][1] = f[x][2] = 0;
	int x1,x2,res,r,son;
	x1 = x2 = -INF; son = 0;
	for (int i = 0; i < v[x].size(); i++) {
		int to = v[x][i];
		if (to == fa) continue;
		++son;
		f[x][0] += (r = dfs(to,x));
		res = max(f[to][0],f[to][1]) - r;
		if (res >= x1) {
			x2 = x1;
			x1 = res;
		}
		else if (res > x2) x2 = res;
	}
	if (son > 0) f[x][1] = max(f[x][0] + x1,0LL) + 1;
	if (son > 1) f[x][2] = max(f[x][0] + x1 + x2,0LL) + 2;
	return max(f[x][0],max(f[x][1],f[x][2]));
}

int main()
{
	#ifdef DMC
		freopen("DMC.txt","r",stdin);
    #else
		freopen("travel.in","r",stdin);
		freopen("travel.out","w",stdout);
	#endif
	
	cin >> n >> x >> y;
	for (int i = 1; i < n; i++) {
		int a,b; scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
		++du[a]; ++du[b];
	}
	if (x >= y) {
		ans = 1LL*(n-1)*y;
		bool flag = 0;
		for (int i = 1; i <= n; i++)
			if (du[i] == n - 1) flag = 1;
		if (flag) ans += (x - y);
		cout << ans;
	}
	else {
		int Max = dfs(1,0);
		ans = 1LL*Max*x + 1LL*(n - 1 - Max)*y;
		cout << ans;
	}
	return 0;
}

T2： 待填

T3：

f[i]:状态为i的点构成的合法树的个数

g[i]:状态为i的构成的合法森林的个

f[i] = ∑g[o-做根的节点]


森林:森林+一棵新树


g[i] = ∑f[op]*g[o-op]  op一定是含o最左边数位的节点

初值:g[0] = 1

一开始以为L、R是儿子数。。傻逼了

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

typedef long long LL;
const int maxn = 20;
const int maxm = 1<<15;
const LL mo = 1E9 + 7;

int n,L[maxn],R[maxn];
LL f[maxm],g[maxm];

int cal(int o)
{
	int ret = 0;
	for (; o; o >>= 1)
		if (o & 1) ++ ret;
	return ret;
}

int main()
{
	#ifdef DMC
		freopen("DMC.txt","r",stdin);
    #else
		freopen("tree.in","r",stdin);
		freopen("tree.out","w",stdout);
	#endif
	
	cin >> n;
	for (int i = 1; i <= n; i++) scanf("%d%d",&L[i],&R[i]);
	g[0] = 1;
	for (int o = 1; o < (1< 1) g[o] = f[o];
		for (int op = (opt - 1) & opt; op; op = (op - 1) & opt)
			g[o] = (g[o] + f[(1<