「杂题记录」彩色挂饰

📅 Last Updated: 2021-01-19

📊 Word Count: 281

⏱️ Estimated Reading Time: 2 minutes

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一道 dp + 图论 + 状压好题

「杂题记录」彩色挂饰

题意简述

给定一张 $n$ 个点 $m$ 条边的无向图，有一些点有颜色，定义颜色相同的点组成的连通块为同色连通块。需要对没有颜色的点进行染色，最小化同色连通块的数量。

无向图满足每个点双的大小 $\le s$ ，给出的颜色分别从为 $1$ 到 $k$ 编号。

$n\le 10^5, 2<k\le 20,s\le 6, n-1\le m < ns$

对于 $10$ % 的数据， $s = 2$ .

分析

$s=2$ 是一档很有启发性的部分分，可以感受到这就是一个树。

不妨设根节点编号为 $1$ 直接树形 dp，设 $f(u, c)$ 为把以 $u$ 为根节点的子树， $u$ 染成颜色 $c$ ，其余的任意染，最小的同色连通块数量。 $f(u, c) =1+\sum_{(u, v) \in E}\left(f(v, c) - 1\right)$ 一个图怎么做？考虑用圆方树将无向图转化为树，然后进行树形 dp。 $f(u, c)$ 的定义转化为以 $u$ 为根的子树，圆点或 方点的父亲染成的颜色。感觉是一类圆方树上 dp 的标准套路。对于圆点： $f(u, c) =1+\sum_{(u, v) \in E}\left(f(v, c) - 1\right)$ 对于方点: 简单dp 一下就可以了。注意到 $s\le 6$ ，瞎搞一下就可以了。注意以下的集合均指每个将一个点双上的点编号离散后的编号点集。

设：定义在点上的函数 $A(x)$ 为与 $x$ 相连的点集。

设：定义在集合上的函数 $C(S)$ ，表示点集 $S$ 是否连通。 $\begin{split} C(\varnothing) &= 1\\ C(S) &= \bigvee_{x \in S} \left( C(S \setminus \{x\}) \land (A(x) \cap S \ne \varnothing) \right) \end{split}$

设：函数 $G(S, c)$ 为将连通块 $S$ 涂成 $c$ ，最小代价。每个连通块对应的子树也算入答案。 $G(S, c)= \begin{cases} \operatorname{INF}, & C(S)=0 \\ 1+\sum_{x\in S}\limits{(f(x, c)-1)} \end{cases}$ 设：函数 $H(S)$ 为将连通块 $S$ 涂上任意一种相同的颜色，最小代价。 $H(S) = \min_{c=1}^{k}\limits{G(S, c)}.$ 设：函数 $L(S)$ 为将连通块 $S$ 切成若干块，每一块涂上相同的颜色，最小代价。 $L(S) = \min\{H(S),\ \min_{s \subseteq S} L(s)+L(S / s)\}$ 就可以求出 $f(u, c)= \min_{S, u \in S}\limits{G(S, c) + L(S_{补})}$ 转移只需要 80 行就写完了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
#include <algorithm>
#include <stack>
#include <cassert>
#include <set>

using namespace std;

const int _ = 4e5 + 100;
const int _K = 30;

int Col[_];
int n, m, k, s;
int head[_];
struct edges{ int node, nxt; } edge[_ << 1]; int tot = 0;
void add(int u, int v){ tot++; edge[tot].node = v; edge[tot].nxt  = head[u]; head[u] = tot; }

vector<int> G[_];
set<int> GG[_];

int dfn[_], low[_], dfc = 0, tmp;
int cnt;
stack<int> S;
void tarjan(int now) {
    dfn[now] = low[now] = ++dfc; S.push(now);
    for(int i = 0; i < (int)G[now].size(); i++) {
        int ex = G[now][i];
        if(!dfn[ex]) {
            tarjan(ex);
            low[now] = min(low[now], low[ex]);
            if(dfn[now] == low[ex]) {
                ++cnt; 
                add(now, cnt); add(cnt, now);
                do add(cnt, tmp = S.top()), add(tmp, cnt), S.pop(); while(tmp != ex);
            }
        } else low[now] = min(low[now], dfn[ex]);
    }
}
int popcnt(int S) { int ans = 0; while(S) ans += ((S & 1) != 0), S >>= 1; return S;  }
int F[_][_K];
const int _S = 15;
int ver = 0;
int _INT_MAX_;
void d(int now, int fa){
    for(int i = head[now]; i; i = edge[i].nxt) {
        int ex = edge[i].node; if(ex == fa) continue;
        d(ex, now);
    }
    if(now <= n) {
        for(int i = 1; i <= k; i++) {
            if(Col[now]) if(Col[now] != i) { F[now][i] = _INT_MAX_; continue; }
            int &ans = F[now][i] = 1;
            for(int j = head[now]; j; j = edge[j].nxt) {
                if(edge[j].node == fa) continue;
                ans += F[edge[j].node][i] - 1;
            }
        }
    } else {
        static int Link[_S], iLink[_], NodeCnt; NodeCnt = 0; 
        for(int i = head[now]; i; i = edge[i].nxt) {
            int ex = edge[i].node;
            Link[++NodeCnt] = ex;
            iLink[ex] = NodeCnt;
        }
        static int A[_S]; static bool C[1 << _S];
        for(int i = 1; i <= NodeCnt; i++) {
            int &ans = A[i] = 0;
            for(int j = 1; j <= NodeCnt; j++){ if(i == j) continue;
                int ex = Link[j]; if(GG[Link[i]].find(ex) == GG[Link[i]].end()) continue;
                ans |= (1 << (j - 1));
            }
        }
        for(int i = 0; i < (1 << NodeCnt); i++) C[i] = 0;
        C[0] = 1; // C[S] 重标号后的点集 S 是否连通. 
        for(int i = 1; i <= NodeCnt; i++) C[1 << (i - 1)] = 1;
        for(int i = 1; i < (1 << NodeCnt); i++){
            bool &ans = C[i];
            for(int j = 1; j <= NodeCnt; j++){
                if(i & (1 << (j - 1))) ; else continue;
                ans = ( ans || (C[i ^ (1 << (j - 1))] && ((A[j] & i) != 0)) );
            }
        }
        static int G[1 << _S][_K]; // G[S][c] 把联通块 S 涂上颜色 c. 需要的次数.
        for(int i = 1; i < (1 << NodeCnt); i++){
            for(int j = 1; j <= k; j++){
                int &ans = G[i][j]; 
                if(!C[i]) { ans = _INT_MAX_; continue; }
                ans = 1;
                for(int l = 1; l <= NodeCnt; l++){
                    if(i & (1 << (l - 1))) if(l != iLink[fa]) ans += F[Link[l]][j] - 1;// , assert(F[Link[l]][j] > 0);
                }
            }
        }
        static int H[1 << _S]; // H[S] 把 连通块 S 涂上同一种颜色 所需要的 最小代价。 
        H[0] = 0; 
        for(int i = 1; i < (1 << NodeCnt); i++){
            int &ans = H[i] = _INT_MAX_;
            for(int j = 1; j <= k; j++) ans = min(ans, G[i][j]);
        }
        static int L[1 << _S]; L[0] = 0; // L[S] 把 连通块 S 分成若干块，每一块分别涂上相同的颜色 最小代价。 
        for(int i = 1; i < (1 << NodeCnt); i++){
            int &ans = L[i] = H[i];
            for(int S0 = i; S0; S0 = (S0 - 1) & i){
                ans = min(ans, L[S0] + L[i ^ S0]);
            }
        }
        for(int i = 1; i <= k; i++){
            int &ans = F[now][i] = _INT_MAX_;
            if(Col[fa]) if(Col[fa] != i) continue;
            int S = (1 << (NodeCnt)) - 1; S ^= (1 << (iLink[fa] - 1));
            ans = min(ans, G[(1 << (iLink[fa] - 1))][i] + L[S]);
            for(int j = S; j ; j = (j - 1) & S){
                ans = min(ans, G[j | (1 << (iLink[fa] - 1))][i] + L[S ^ j]);
            }
        }
    }
}

int main(){ 
    // freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false);
    cin >> n >> m >> k >> s; cnt = n; _INT_MAX_ = 3 * n;
    for(int i = 1; i <= n; i++) cin >> Col[i];
    for(int i = 1; i <= m; i++){
        int u, v; cin >> u >> v;
        G[u].push_back(v); GG[u].insert(v);
        G[v].push_back(u); GG[v].insert(u);
    }
    tarjan(1);
    d(1, 1);
    int ans = _INT_MAX_;
    for(int i = 1; i <= k; i++) ans = min(ans, F[1][i]);
    printf("%d", ans);
    return 0;
}

📚 Version History

Source Code Change History

v1 📅 2021-01-19 15:35:06

🔗 Source Hash: 34e1f3705a1f0e1a4755e793665b424f5cef6901d4da5e2752a40ded97ff3309

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    pdftitle={「杂题记录」彩色挂饰},
    pdfauthor={Jiayi Su (ShuYuMo)},
    hidelinks,
    pdfcreator={LaTeX via pandoc}}
  
  \title{「杂题记录」彩色挂饰}
  \author{Jiayi Su (ShuYuMo)}
  \date{2021-01-19 15:35:06}
  
  \begin{document}
  \maketitle
  
  \setstretch{1.3}
  一道 dp + 图论 + 状压好题
  
  \section{「杂题记录」彩色挂饰}\label{ux6742ux9898ux8bb0ux5f55ux5f69ux8272ux6302ux9970}
  
  \subsection{题意简述}\label{ux9898ux610fux7b80ux8ff0}
  
  给定一张 \(n\) 个点 \(m\)
  条边的无向图，有一些点有颜色，定义颜色相同的点组成的连通块为同色连通块。需要对没有颜色的点进行染色，最小化同色连通块的数量。
  
  无向图满足每个点双的大小 \(\le s\) ，给出的颜色分别从为 \(1\) 到 \(k\)
  编号。
  
  \(n\le 10^5, 2 patch level 1
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  (.../xeCJK.sty
  (.../ctexhook.sty)
  (.../xtemplate.sty)
  (.../xeCJK.cfg))
  (.../upquote.sty
  (.../textcomp.sty))
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  (.../microtype-xetex.def)
  (.../microtype.cfg))
  (.../setspace.sty)
  (.../parskip.sty
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  (.../ltxcmds.sty)
  (.../kvsetkeys.sty)))
  (.../fancyvrb.sty)
  (.../bookmark.sty
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  (.../pdfescape.sty
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  (.../hycolor.sty)
  (.../auxhook.sty)
  (.../nameref.sty
  (.../refcount.sty)
  (.../gettitlestring.sty))
  (.../pd1enc.def)
  (.../intcalc.sty)
  (.../puenc.def)
  (.../url.sty)
  (.../bitset.sty
  (.../bigintcalc.sty))
  (.../atbegshi-ltx.sty))
  (.../hxetex.def
  (.../stringenc.sty)
  (.../rerunfilecheck.sty
  (.../atveryend-ltx.sty)
  (.../uniquecounter.sty)))
  (.../bkm-dvipdfm.def))
  (.../xurl.sty)
  (.../input.aux)
  *geometry* driver: auto-detecting
  *geometry* detected driver: xetex
  (.../mt-LatinModernRoman.cfg)
  (.../omllmm.fd)
  (.../umsa.fd)
  (.../mt-msa.cfg)
  (.../umsb.fd)
  (.../mt-msb.cfg)
  
  LaTeX Font Warning: Font shape `TU/ARPLUKaiCN(0)/b/n' undefined
  (Font)              using `TU/ARPLUKaiCN(0)/m/n' instead on input line 116.
  
  [1]
  Missing character: There is no 补 (U+8865) in font LatinModernMath-Regular/OT:sc
  ript=math;language=dflt;+ssty=0;!
  Missing character: There is no 补 (U+8865) in font LatinModernMath-Regular/OT:sc
  ript=math;language=dflt;+ssty=0;!
  
  Package xeCJK Warning: Unknown CJK family `\CJKttdefault' is being ignored.
  (xeCJK)                
  (xeCJK)                Try to use `\setCJKmonofont[<...>]{<...>}' to define
  (xeCJK)                it.
  
  [2]
  
  LaTeX Font Warning: Font shape `TU/ARPLUKaiCN(0)/m/it' undefined
  (Font)              using `TU/ARPLUKaiCN(0)/m/n' instead on input line 250.
  
  [3] [4] [5] (.../input.aux)
  
  LaTeX Font Warning: Some font shapes were not available, defaults substituted.
  
   )
  Output written on .../input.pdf (5 pages).
  Transcript written on .../input.log.
[WARNING] Missing character: There is no 补 (U+8865) (U+8865) in font LatinModernMath-Regular/OT:sc
[WARNING] Missing character: There is no 补 (U+8865) (U+8865) in font LatinModernMath-Regular/OT:sc