Problem

每个点都可以选择降落士兵,然后当一个点的子节点被攻占的数量超过读入中的限制后,这个城市也被占领。
每个点降落士兵都有一定的代价,问把这一个图全部攻占的最小代价。

Solution

这显然和儿子有关还与父亲有关
我们假设f[x]表示x在父亲之前被攻占,g[x]表示x再父亲之后被攻占
显然有f[x]>=g[x]
在x放兵时,f[x]=g[x]=p[x]+∑gv
当x不放兵时,显然是在其儿子里选c[x]个取f[x](计算g[x]时为c[x]-1),剩下的取g[x]。排序即可。

Notice

感觉不是很好理解

Code

#include<cmath>
#include<vector>
#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 = 5e4;
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');}
vector<int> edge[N + 5];
ll Val[N + 5], f[N + 5], g[N + 5];
int Limit[N + 5];
int cmp(int x, int y)
{
	return f[x] - g[x] < f[y] - g[y];
}
void dfs(int u, int fa)
{
	for (auto i = edge[u].begin(); i != edge[u].end();  i++)
		if (*i == fa) i = edge[u].erase(i) - 1;
		else dfs(*i, u);
}
void dp(int u)
{
	ll T = 0;
	for (auto i : edge[u]) dp(i), T += g[i];
	f[u] = g[u] = Val[u] + T;
	sort(edge[u].begin(), edge[u].end(), cmp);
	int p = edge[u].size();
	rep(i, 0, min(p, Limit[u] - 1) - 1) T += f[edge[u][i]] - g[edge[u][i]];
	if (Limit[u] - 1 <= edge[u].size()) g[u] = min(g[u], T);
	if (Limit[u] <= edge[u].size()) f[u] = min(f[u], T + f[edge[u][Limit[u] - 1]] - g[edge[u][Limit[u] - 1]]);
}
int sqz()
{
	int H_H = read();
	while (H_H--)
	{
		memset(f, 0, sizeof f);
		memset(g, 0, sizeof g);
		int n = read();
		rep(i, 1, n) edge[i].clear();
		rep(i, 1, n - 1)
		{
			int u = read(), v = read();
			edge[u].push_back(v);
			edge[v].push_back(u);
		}
		rep(i, 1, n) Val[i] = read();
		rep(i, 1, n) Limit[i] = read();
		dfs(1, 0);
		dp(1);
		write(f[1]); puts("");
	}
	return 0;
}
posted on 2017-10-30 20:42  WizardCowboy  阅读(93)  评论(0编辑  收藏  举报