[51nod1789] 跑得比谁都快

题面

题解

设\(f[i]\)为根节点到\(i\)的最小耗时

设\(S\)为\(i\)的祖先集合, 可以得到
\[ f[i] = min(f[j] + (i - j)^p),j \in S \]
对于\((i - j)^p\), 我们有
\[ ((i + 1) - (j + 1))^p + (i - j)^p \geq ((i + 1) - j)^p + (i - (j + 1))^p \]
可以发现这是一个满足四边形不等式的式子

直接上决策单调性即可(我这个写法是看的别人的, 应该是对的吧)

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define itn int
#define reaD read
#define N 100005
using namespace std;

int n, p, w[N], cnt; 
long long pw[N], ans; 

template < typename T > 
inline T read()
{
    T x = 0, w = 1; char c = getchar();
    while(c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * w;
}

namespace Graph
{
    int head[N];
    struct edge { int to, next; } e[N]; 
    inline void adde(int u, int v) { e[++cnt] = (edge) { v, head[u] }; head[u] = cnt; }
};

using namespace :: Graph; 

long long fpow(long long x, int y = p)
{
    long long res = 1;
    for( ; y; y >>= 1, x = 1ll * x * x)
        if(y & 1) res = 1ll * res * x;
    return res; 
}

namespace DFS
{
    long long f[N];
    int top, stk[N], pos[N]; 
    struct node { int l, r, id; } q[N]; 
    void dfs(int u, int fa)
    {
        if(u == 1) stk[++top] = u, f[u] = 0, pos[u] = top;
        else
        {
            int num; 
            long long tmp = f[0]; 
            for(int i = pos[fa]; i <= top; i++)
            {
                long long res = f[stk[i]] + w[stk[i]] + fpow(u - stk[i], p);
                if(res <= tmp) num = i, tmp = res; 
            }
            f[u] = tmp;
            pos[u] = num;
            stk[++top] = u; 
        }
        bool flag = 0; 
        for(int i = head[u]; i; i = e[i].next)
            flag = 1, dfs(e[i].to, u); 
        if(!flag) ans = min(ans, f[u]); 
        top--; 
    }
}; 

using namespace :: DFS; 

int main()
{
    n = read <int> (); p = read <int> ();
    for(int i = 1; i <= n; i++)
    {
        w[i] = read <int> (); int u = read <int> ();
        if(u) adde(u, i); 
    }
    memset(f, 0x3f, sizeof(f)); 
    ans = f[0]; 
    dfs(1, 0);
    printf("%lld\n", ans); 
    return 0;
}