CodeForces 804C Ice cream coloring

题解：

这个题目中最关键的一句话是，把任意一种类型的冰激凌所在的所有节点拿下来之后，这些节点是一个连通图(树)。

所以就不会存在多个set+起来之后是一个新的完全图。

所以只要直接去做就好了。

对于每个节点来说，染色。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 3e5 + 100;
vector<int> vc[N], e[N];
int col[N];
int vis[N];
int mx = 0;
void dfs(int o, int u){
    for(int v : vc[u]){
        vis[col[v]]++;
    }
    int b = 1;
    for(int v : vc[u]){
        if(col[v]) continue;
        while(vis[b]) ++b;
        vis[b] = 1;
        col[v] = b;
    }
    mx = max(mx, b);
    for(int v : vc[u]){
        vis[col[v]] = 0;
    }
    for(int v : e[u]){
        if(v == o) continue;
        dfs(u, v);
    }
}
int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i){
        int si, tv;
        scanf("%d", &si);
        for(int j = 1; j <= si; ++j){
            scanf("%d", &tv);
            vc[i].pb(tv);
        }
    }
    int u, v;
    for(int i = 1; i < n; ++i){
        scanf("%d%d", &u, &v);
        e[u].pb(v); e[v].pb(u);
    }
    dfs(0, 1);
    printf("%d\n", mx);
    for(int i = 1; i <= m; ++i){
        if(!col[i]) col[i] = 1;
        printf("%d%c", col[i], " \n"[i==m]);
    }
    return 0;
}

View Code

CodeForces 804C Ice cream coloring

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