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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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; }