树形dp Codeforces Round #168 (Div. 2) D题 Zero Tree

Zero Tree

A tree is a graph with n vertices and exactly n - 1 edges; this graph should meet the following condition: there exists exactly one shortest (by number of edges) path between any pair of its vertices.

A subtree of a tree T is a tree with both vertices and edges as subsets of vertices and edges of T.

You’re given a tree with n vertices. Consider its vertices numbered with integers from 1 to n. Additionally an integer is written on every vertex of this tree. Initially the integer written on the i-th vertex is equal to vi. In one move you can apply the following operation:

Select the subtree of the given tree that includes the vertex with number 1.
Increase (or decrease) by one all the integers which are written on the vertices of that subtree.
Calculate the minimum number of moves that is required to make all the integers written on the vertices of the given tree equal to zero.

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题目大意：给一颗 n 个结点的树，每个结点都有一个权值，选择一颗包含结点 1 的子树（1是标号不是权值），对这颗子树上的权值减去或者加上 1 ，问最少操作次数可以使树上的每个点权值变为0；

分析可知，对于一颗子树的根结点，最少要加上其子节点的权值的最小值的绝对值（如果子结点权值为负），最少要减去其子节点的权值最大值（如果子结点权值为正）；

所以可以开两个数组，decr[i] 表示 i 为根的子树要减去的值，inc[i] 表示 i 为根的子树要加上的值；每次取最大值即可；

注意更新根结点的操作；

代码：

#include
#define LL long long
#define pa pair
#define ls k<<1
#define rs k<<1|1
#define inf 0x3f3f3f3f
using namespace std;
const int N=100100;
const int M=2000100;
const LL mod=1e9+7;
int head[N],cnt,n; 
LL inc[N],decr[N],a[N];
struct Node{
     
	int to,nex;
}edge[N*2];
void add(int p,int q){
     
	edge[cnt].to=q,edge[cnt].nex=head[p],head[p]=cnt++;
}
void dfs(int p,int ft){
     
	LL mx1=-1,mx2=-1;
	for(int i=head[p];~i;i=edge[i].nex){
     
		int q=edge[i].to;
		if(q!=ft){
     
			dfs(q,p);
			mx1=max(mx1,inc[q]),mx2=max(mx2,decr[q]);
		}
	}
	if(mx1==-1&&mx2==-1){
     
		if(a[p]>=0) decr[p]=a[p];
		else inc[p]=-a[p];
	}
	else{
     
		decr[p]=mx2,inc[p]=mx1; 
		a[p]+=mx1-mx2;
		if(a[p]>=0) decr[p]+=a[p];
		else inc[p]+=-a[p];
	}
}
int main(){
     
	memset(head,-1,sizeof(head));
	scanf("%d",&n);
	for(int i=1;i<n;i++){
     
		int a,b;scanf("%d%d",&a,&b);
		add(a,b),add(b,a);
	}
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	dfs(1,-1);
	printf("%lld\n",decr[1]+inc[1]);
	return 0;
}