Codeforces Round #646 (Div. 2)E. Tree Shuffling 题解（dfs）

题目链接

题目大意

给你一颗树，每一个节点有一个a[i],b[i],c[i]值，你要把b[i]变成c[i]，b[i]和c[i]为[0,1]，你操作的方法是，选择一个节点x，选择他的k个子树，然后进行交换他们的b[i]值，使满足要求，每次操作的cost为a[x]*k，求满足题目要求所需要的最小cost，如果不能满足要求，则输出-1.

题目思路

基本思路想到了，但是不会实现，码力太差，直接放标准解析。

Key Idea: Let the parent of node i be p. If a[i]≥a[p], we can do the shuffling which was done at i, at p instead. Thus, we can do the operation a[i]=min(a[i],a[p]).

Detailed Explanation: Let us denote nodes that have bi=1 and ci=0 as type 1, and those that have bi=0 and ci=1 as type 2. Firstly, the answer is −1 if and only if the number of nodes of type 1 and type 2 are unequal.

We also observe that only nodes of type 1 and 2 should be shuffled - it is unoptimal to shuffle those which already have b[i]=c[i]. Thus, we should try to exchange the values of type 1 and type 2 nodes.

We use the key idea by going down from the root, and at every node i, setting a[i]=min(a[i],a[p]) where p is the parent node of i in the tree. Thus, the a[i]'s now follow a special structure: they are non-increasing from the root to the leaves!

This paves the way for our greedy solution: we will go upwards from the leaves, and at each node i interchange type 1 and type 2 nodes until we have no nodes in one of these types. Then, we pass on the remaining nodes to the parent to be shuffled.

代码

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
int n,a[maxn],b[maxn],c[maxn],head[maxn],cnt;
ll cost;
struct node{int to,next;
}e[maxn<<1];
void add(int u,int v){e[++cnt].to=v;e[cnt].next=head[u];head[u]=cnt;
}
pair<int,int> dfs(int son,int fa,int mi){pair<int,int> x;x.first=x.second=0;if(b[son]==1&&c[son]==0){x.first++;}else if(b[son]==0&&c[son]==1){x.second++;}for(int i=head[son];i;i=e[i].next){if(e[i].to!=fa){pair<int,int> y=dfs(e[i].to,son,min(mi,a[son]));x.first+=y.first;x.second+=y.second;}}if(a[son]<=mi){int take=min(x.first,x.second);cost+=1ll*2*take*a[son];x.first-=take;x.second-=take;}return x;
}
int main(){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%d %d %d",&a[i],&b[i],&c[i]);}for(int i=1,u,v;i<=n-1;i++){scanf("%d %d",&u,&v);add(u,v),add(v,u);}pair<int,int> ans=dfs(1,1,a[1]);if(ans.first==0&&ans.second==0){printf("%lld\n",cost);}else{printf("-1\n");}return 0;
}