[USACO17DEC] Barn Painting

最新推荐文章于 2021-11-22 18:56:55 发布

转载最新推荐文章于 2021-11-22 18:56:55 发布 · 79 阅读

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原文链接：http://www.cnblogs.com/JYYHH/p/8452096.html

本文探讨了在大型农场中，对于N个谷仓使用三种颜色进行染色的问题。目标是在确保直接相连的谷仓不使用相同颜色的前提下，计算所有合法染色方案的数量。文章提供了一种基于树形DP的方法，并附带了完整的代码实现。

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题目描述

Farmer John has a large farm with $\le N \le 10^51≤N≤105 ), some of which are already painted and some not yet painted. Farmer John wants to paint these remaining barns so that all the barns are painted, but he only has three paint colors available. Moreover, his prize cow Bessie becomes confused if two barns that are directly reachable from one another are the same color, so he wants to make sure this situation does not happen.$

It is guaranteed that the connections between the

How many ways can Farmer John paint the remaining yet-uncolored barns?

输入输出格式

输入格式：

The first line contains two integers $\le K \le N0≤K≤N ), respectively the number of barns on the farm and the number of barns that have already been painted.$

The next $\le x, y \le N, x \neq y1≤x,y≤N,x≠y ) describing a path directly connecting barns xx and yy .$

The next $\le b \le N1≤b≤N , 1 \le c \le 31≤c≤3 ) indicating that barn bb is painted with color cc .$

输出格式：

Compute the number of valid ways to paint the remaining barns, modulo $10^9 + 7109+7 , such that no two barns which are directly connected are the same color.$

输入输出样例

输入样例#1：

4 1
1 2
1 3
1 4
4 3

输出样例#1：

8


树上dp求相邻节点不同的染色方案。
已染色的就把该节点的其他颜色的方案数置为0即可。
(这个难度评价有毒，把我骗进来了hhhh)

#include<bits/stdc++.h>
#define ll long long
#define maxn 100005
#define pb push_back
using namespace std;
const int ha=1000000007;
vector<int> g[maxn];
int f[maxn][4],col[maxn];
int n,m,k,ans;

inline int add(int x,int y){
    x+=y;
    if(x>=ha) return x-ha;
    else return x;
}

void dfs(int x,int fa){
    f[x][1]=f[x][2]=f[x][3]=1;
    
    int to,tmp;
    for(int i=g[x].size()-1;i>=0;i--){
        to=g[x][i];
        if(to==fa) continue;
        
        dfs(to,x);
        
        tmp=add(f[to][1],add(f[to][2],f[to][3]));
        for(int j=1;j<=3;j++) f[x][j]=f[x][j]*(ll)add(tmp,ha-f[to][j])%ha;
    }
    
    if(col[x]){
        for(int i=1;i<=3;i++) if(i!=col[x]) f[x][i]=0;
    }
}

int main(){
    int uu,vv;
    scanf("%d%d",&n,&k);
    for(int i=1;i<n;i++){
        scanf("%d%d",&uu,&vv);
        g[uu].pb(vv),g[vv].pb(uu);
    }
    for(int i=1;i<=k;i++){
        scanf("%d%d",&uu,&vv);
        col[uu]=vv;
    }
    
    dfs(1,0);
    
    ans=add(add(f[1][2],f[1][1]),f[1][3]);
    printf("%d\n",ans);
    
    return 0;
}

转载于:https://www.cnblogs.com/JYYHH/p/8452096.html