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P5024 [NOIP2018 提高组] 保卫王国

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思路

如果没有强制,那就是一个简单的树形DP,我们用 \(f[i][0/1]\) 表示 \(i\) 的子树内,\(i\) 选或不选的最小代价;用 \(g[i][0/1]\) 表示整个树减去 \(i\) 的子树,\(i\) 选或不选单最小代价。这类似于换根DP

有了强制,说明我们的DP有一些状态不可取,虽然我们不能退回去再做一次DP,但我们可以记录一些信息,达到 \(\log n\) 来快速计算

我们设 \(h[i][j][0/1][0/1]\) 代表 \(i\) 结点和向上跳 \(2^j\) 个结点到某个祖先,然后 \(i\) 选或不选,祖先选或不选;里面存的是不包含 \(f[i]\) 和 \(g[fa[i][j]]\) 的最小代价(也就是整棵树,除去“ \(i\) 的子树”和“不含 \(fa[i][j]\) 的子树的树”)

每次强制要求 \(u,v\),我们就用倍增求 \(\text{LCA}(u,v)\) 的同时,将路径上的 \(h\) 合并:

  1. 当 \(v\) 是 \(u\) 的祖先时,那么答案就是 \(h[u][v][a][b] + f[u][a]+g[v][b]\)(\(a,b\) 表示 \(u,v\) 强制选或不选,这里的 \(h\) 是倍增合并后的)

  2. 如果不是祖孙关系,那么就一路合并到 \(\text{LCA}(u,v)\) 的儿子,最后分类讨论两个儿子的状态,取最小值


代码

#include<iostream>
#include<fstream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#define LL long long
#define FOR(i, x, y) for(int i = (x); i <= (y); i++)
#define ROF(i, x, y) for(int i = (x); i >= (y); i--)
#define PFOR(i, x) for(int i = he[x]; i; i = r[i].nxt)
inline int reads()
{
    int sign = 1, re = 0; char c = getchar();
    while(c < '0' || c > '9'){if(c == '-') sign = -1; c = getchar();}
    while('0' <= c && c <= '9'){re = re * 10 + (c - '0'); c = getchar();}
    return sign * re;
}
const LL INF = 1e18;
int n, m, a[100005]; char _[2];
struct Node
{
    int to, nxt;
}r[200005]; int he[100005];
inline void Edge_add(int u, int v)
{
    static int cnt = 0;
    r[++cnt] = (Node){v, he[u]};
    he[u] = cnt;
}
LL f[100005][2], g[100005][2], h[100005][17][2][2];
int fa[100005][17], dep[100005];
void dfs1(int now)
{
    f[now][1] = a[now];
    PFOR(i, now)
    {
        int to = r[i].to;
        if(to == fa[now][0]) continue;
        fa[to][0] = now, dep[to] = dep[now] + 1;
        dfs1(to);
        f[now][0] += f[to][1], f[now][1] += std::min(f[to][0], f[to][1]);
    }
}
void dfs2(int now)
{
    PFOR(i, now)
    {
        int to = r[i].to;
        if(to == fa[now][0]) continue;
        g[to][0] = g[now][1] + (f[now][1] - std::min(f[to][0], f[to][1]));
        g[to][1] = std::min(g[to][0], g[now][0] + (f[now][0] - f[to][1]));
        dfs2(to);
    }
}
inline void Init()
{
    FOR(i, 2, n)
    {
        h[i][0][0][0] = INF;
        h[i][0][0][1] = h[i][0][1][1] = f[fa[i][0]][1] - std::min(f[i][0], f[i][1]);
        h[i][0][1][0] = f[fa[i][0]][0] - f[i][1];
    }
    FOR(j, 1, 16)
    {
        FOR(i, 2, n)
        {
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
            if(!fa[i][j]) continue;
            FOR(a, 0, 1) FOR(b, 0, 1)
            {
                h[i][j][a][b] = INF;
                FOR(c, 0, 1)
                    h[i][j][a][b] = std::min(h[i][j][a][b], h[i][j - 1][a][c] + h[fa[i][j - 1]][j - 1][c][b]);
            }
        }
    }
}
inline LL work(int u, int x, int v, int y)
{
    LL lu[2] = {INF, INF}, nu[2] = {INF, INF};
    LL lv[2] = {INF, INF}, nv[2] = {INF, INF};
    lu[x] = f[u][x];
    ROF(i, 16, 0)
        if(dep[fa[u][i]] >= dep[v])
        {
            FOR(j, 0, 1) FOR(k, 0, 1)
                nu[k] = std::min(nu[k], lu[j] + h[u][i][j][k]);
            lu[0] = nu[0], lu[1] = nu[1], nu[0] = nu[1] = INF,
            u = fa[u][i];
        }
    if(u == v) return lu[y] + g[v][y];
    lv[y] = f[v][y];
    ROF(i, 16, 0)
        if(fa[u][i] != fa[v][i])
        {
            FOR(j, 0, 1) FOR(k, 0, 1)
                nu[k] = std::min(nu[k], lu[j] + h[u][i][j][k]),
                nv[k] = std::min(nv[k], lv[j] + h[v][i][j][k]);
            lu[0] = nu[0], lu[1] = nu[1], nu[0] = nu[1] = INF, u = fa[u][i];
            lv[0] = nv[0], lv[1] = nv[1], nv[0] = nv[1] = INF, v = fa[v][i];
        }
    int lca = fa[u][0];
    LL re[2];
    re[0] = f[lca][0] + g[lca][0] - f[u][1] - f[v][1] + lu[1] + lv[1];
    re[1] = f[lca][1] + g[lca][1] - std::min(f[u][0], f[u][1]) - std::min(f[v][0], f[v][1]) + std::min(lu[0], lu[1]) + std::min(lv[0], lv[1]);
    return std::min(re[0], re[1]);
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("test.in", "r", stdin);
    freopen("test.out", "w", stdout);
#endif
    n = reads(), m = reads(); scanf("%s", &_);
    FOR(i, 1, n) a[i] = reads();
    FOR(i, 1, n - 1)
    {
        int u = reads(), v = reads();
        Edge_add(u, v), Edge_add(v, u);
    }
    dep[1] = 1, dfs1(1), dfs2(1), Init();
    FOR(i, 1, m)
    {
        int u = reads(), x = reads(), v = reads(), y = reads();
        if(dep[u] < dep[v]) std::swap(u, v), std::swap(x, y);
        if(!x && !y && fa[u][0] == v) {puts("-1"); continue;}
        printf("%lld\n", work(u, x, v, y));
    }
    return 0;
}

标签:std,NOIP2018,fa,reads,lu,include,P5024,nu,保卫
来源: https://www.cnblogs.com/zuytong/p/16589447.html