Gym 101964C

思维不够,好气

Gym 101964C

题目描述

给你一棵树,询问你黑点最长路径最短的值是多少

题解

二分答案

定义 $dp[i][j]$ 表示以i为根的子树,选出来的黑点中距 $i$ 节点距离不会超过 $j$,所能选出最多的黑点个数.
并记 $lim=\min(M−1−j,j−1)$
那么转移就是:
假设 $v_1,v_2,…,v_m$ 是 $i$ 的儿子节点.

$\min _{1\leq p\leq m}{dp[vp][j−1]+\sum {q!=p} dp[v_q][lim]}→dp[i][j]$

$dp[i][j−1]→dp[i][j]$

最后只要看 $dp[1][M]≥k$.

AC Code

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//#define NOSTDCPP
//#define Cpp11
//#define Linux_System
#ifndef NOSTDCPP
#include <bits/stdc++.h>
#else
#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <complex>
#include <cstring>
#include <cstdio>
#include <deque>
#include <exception>
#include <functional>
#include <iomanip>
#include <iostream>
#include <istream>
#include <iterator>
#include <list>
#include <map>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>
#endif
# ifdef Linux_System
# define getchar getchar_unlocked
# define putchar putchar_unlocked
# endif
# define RESET(_) memset(_, 0, sizeof(_))
# define RESET_(_, val) memset(_, val, sizeof(_))
# define fi first
# define se second
# define pb push_back
# define midf(x, y) ((x + y) >> 1)
# define DXA(_) (((_) << 1))
# define DXB(_) (((_) << 1) | 1)
# define distance __Chtholly__
# define x1 __Mercury__
# define y1 __bbtl04__
# define index __ikooo__
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector <int> vi;
typedef set <int> si;
typedef pair <int, int> pii;
typedef long double ld;
const int MOD = 1e9 + 7;
const int maxn = 109;
const int maxm = 300009;
const ll inf = 1e18;
const double pi = acos(-1.0);
const double eps = 1e-6;
ll myrand(ll mod){return ((ll)rand() << 32 ^ (ll)rand() << 16 ^ rand()) % mod;}
template <class T>
inline bool scan_d(T & ret)
{
char c;
int sgn;
if(c = getchar(), c == EOF)return false;
while(c != '-' && (c < '0' || c > '9'))c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while(c = getchar(), c >= '0' && c <= '9')
ret = ret * 10 + (c - '0');
ret *= sgn;
return true;
}
#ifdef Cpp11
template <class T, class ... Args>
inline bool scan_d(T & ret, Args & ... args)
{
scan_d(ret);
scan_d(args...);
}
#define cin.tie(0); cin.tie(nullptr);
#define cout.tie(0); cout.tie(nullptr);
#endif
inline bool scan_ch(char &ch)
{
if(ch = getchar(), ch == EOF)return false;
while(ch == ' ' || ch == '\n')ch = getchar();
return true;
}
template <class T>
inline void out_number(T x)
{
if(x < 0)
{
putchar('-');
out_number(- x);
return ;
}
if(x > 9)out_number(x / 10);
putchar(x % 10 + '0');
}
int n, m, A[maxn], dis[maxn][maxn];
vi g[maxn];
void dfs(int u, int fa, int mid)
{
//遍历子树
for(register int i = 0, gsize = g[u].size(), v; i < gsize; ++ i)
if((v = g[u][i]) != fa)
dfs(v, u, mid);
if(A[u] == 1) dis[u][0] = 1;
for(register int i = 1, lim, sum, j, gsize = g[u].size(), v, tmp; i <= mid; ++ i)
{
lim = min(mid - 1 - i, i - 1); //子树的两边进行遍历
sum = 0;
if(lim >= 0)
{
for(j = 0; j < gsize; ++ j) // 从远点转移到当前点的权值和
if((v = g[u][j]) != fa)
sum += dis[v][lim];
}
dis[u][i] = max(dis[u][i], dis[u][i - 1]);
for(j = 0; j < gsize; ++ j) // 从儿子转移到父亲
{
if((v = g[u][j]) == fa) continue;
tmp = dis[v][i - 1];
if(lim >= 0) tmp += sum - dis[v][lim];
dis[u][i] = max(dis[u][i], tmp + A[u]);
}
}
}
bool f(int x)
{
RESET(dis);
dfs(1, 0, x);
for(register int i = 1; i <= n; ++ i)
if(dis[i][x] >= m)
return false;
return true;
}
int main()
{
scanf("%d %d", &n, &m);
for(register int i = 1; i <= n; ++ i) scanf("%d", A + i);
for(register int i = 1, f, t; i < n; ++ i)
{
scanf("%d %d", &f, &t);
g[f].pb(t);
g[t].pb(f);
}
int l = 0, r = n, ans, mid;
while(l <= r)
{
mid = midf(l, r);
if(! f(mid))
{
ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
printf("%d\n", ans);
return 0;
}
/*
6 3
1 1 0 1 1 1
1 2
1 3
1 4
3 5
3 6
9 4
1 0 1 0 1 0 0 1 1
1 2
2 4
2 3
4 5
1 6
6 7
6 8
7 9
10 6
1 0 0 1 1 1 1 1 1 1
1 2
1 3
2 4
2 5
3 6
3 7
6 8
6 9
9 10
*/
文章目录
  1. 1. Gym 101964C
    1. 1.1. 题目描述
    2. 1.2. 题解
    3. 1.3. AC Code
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