标签：scanf ble inpu nim ted ati pos 父节点 get
Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.
The game is played on a tree having n nodes, numbered from 1 to n. Each node i has an initial value initi, which is either 0 or 1. The root of the tree is node 1.
One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node x. Right after someone has picked node x, the value of node x flips, the values of sons of x remain the same, the values of sons of sons of xflips, the values of sons of sons of sons of x remain the same and so on.
The goal of the game is to get each node i to have value goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.
The next line contains n integer numbers, the i-th of them corresponds to initi (initi is either 0 or 1). The following line also contains ninteger numbers, the i-th number corresponds to goali (goali is either 0 or 1).
In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi. cells into walls so that all the remaining cells still formed a connected area. Help him.
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
2
4
7
题意：
给出一棵树，有n个节点，根为1，每个节点的值不是0就是1，每次可以翻转一个节点，翻转后，当前节点值翻转，儿子节点值不变，儿子的儿子翻转，依次类推，求最少翻转次数，使得每个节点的值与目标结果相同。
题解:
由于翻转一个节点，其子节点有的会改变，因此贪心的从根开始搜索，如果当前节点与目标不同，那么一定要翻转，因为只影响子节点，因此其父节点已经到达目标结果将不受影响。
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100005;
int init[maxn],goal[maxn];
vector<int>ans,itv[maxn];
void dfs(int u,int fa,int odd,int even)
{
if (odd) init[u] ^= 1;
if (init[u] != goal[u])
{
odd ^= 1;
ans.push_back(u);
}
int size = itv[u].size();
for (int i = 0;i < size;i++)
{
int v = itv[u][i];
if (fa == v) continue; //树是双向的，因此此处保证从根单向搜索
dfs(v,u,even,odd);
}
}
int main()
{
int n,u,v;
scanf("%d",&n);
for (int i = 1;i < n;i++)
{
scanf("%d%d",&u,&v);
itv[u].push_back(v);
itv[v].push_back(u);
}
for (int i = 1;i <= n;i++) scanf("%d",&init[i]);
for (int i = 1;i <= n;i++) scanf("%d",&goal[i]);
dfs(1,1,0,0);
int size = ans.size();
printf("%d\n",size);
for (int i = 0;i < size;i++) printf("%d\n",ans[i]);
return 0;
}
Codeforces Round #245 (Div. 1)A. Xor-tree（深搜）
标签：scanf ble inpu nim ted ati pos 父节点 get
原文地址：http://www.cnblogs.com/zzy19961112/p/7143539.html
