OpenJudge

7:The xor-longest Path

总时间限制:
2000ms
内存限制:
65536kB
描述

In an edge-weighted tree, the xor-length of a path pis defined as the xor sum of the weights of edges on p:

⊕ is the xor operator.

We say a path the xor-longest path if it has the largest xor-length. Given an edge-weighted tree with n nodes, can you find the xor-longest path?  

输入

The input contains several test cases. The first line of each test case contains an integer n(1<=n<=100000), The following n-1 lines each contains three integers u(0 <= u < n),v(0 <= v < n),w(0 <= w < 2^31), which means there is an edge between node u and v of length w.

输出
For each test case output the xor-length of the xor-longest path.
样例输入
4
0 1 3
1 2 4
1 3 6
样例输出
7
提示

The xor-longest path is 0->1->2, which has length 7 (=3 ⊕ 4)



#include < algorithm >
#include < cstdio >
#include < cstring >
#include < vector >
using namespace std;

const int MAXN = 100010;
const int MAXM = 200010;
struct Trie {
Trie *children[2];
int value;
} root, pool[MAXN << 5], *ptr, emptyNode;
// Trie树是一棵二叉树,左孩子表示当前二进制位为0,右孩子表示当前二进制位为1
// 叶子节点额外保存value,表示当前路径对应的数,如根-左-右-右这个节点的value为(011)2=3

int n;
// 节点个数

int head[MAXN], succeed[MAXM], vertex[MAXM], weight[MAXM], now;
// head,succeed使用链表保存边表;vertex,weight存每条边的终点以及权值

int rootXor[MAXN];
// 每个点到根的路径的xor值

void addEdge(int u, int v, int w) {
// 往链表中添加一条u->v,权值为w的边
________(1)________
// 新的节点的后继为u节点原来的链表头
vertex[now] = v;
// 新的边的终点为v
weight[now] = w;
// 新的边的权值为w
head[u] = now++;
// 指定u节点新的链表头为now
}

void readTree() {
// 读入树
memset(head, -1, sizeof head);
now = 0;
for (int i = 0; i < n - 1; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
addEdge(u, v, w);
addEdge(v, u, w);
}
}

void DFS(int x = 0, int father = -1) {
// DFS求出每个点到0号点的路径的xor值,并填到rootXor数组中
// 参数x表示当前递归到的节点,father表示当前点的父亲
for (int now = head[x]; now != -1; now = succeed[now]) {
// 遍历x节点的链表,找到所有x的出边
int y = vertex[now];
int w = weight[now];
// 从x到y有一条权值为w的边
if (________(2)________)
continue;
// 忽略返回父亲的边
rootXor[y] = ________(3)________;
// 计算rootXor[y]的值,为y到0号点的路径的xor值
DFS(y, x);
}
}

int getNthBit(int value, int nBit) {
return value >> nBit & 1;
}

void insertTrie(Trie *node, int value, int nBit) {
// 往Trie中插入value
// 当前在node这个节点,处理到value的第nBit个bit
// 如果nBit为-1,说明已经处理完毕,node为叶子节点
if (nBit == -1)
node->value = ________(4)________;
else {
bool bit = ________(5)________;
// bit为0说明当前的节点应为node的左孩子
// bit为1说明当前的节点应为node的右孩子
if (!node->children[bit]) {
node->children[bit] = ptr++;
*node->children[bit] = emptyNode;
}
insertTrie(node->children[bit], value, nBit - 1);
}
}

void buildTrie() {
// 将所有rootXor[i]插入Trie中
root = emptyNode;
ptr = pool;
for (int i = 0; i < n; i++)
insertTrie(&root, rootXor[i], 30);
}

int queryTrie(Trie *node, int value, int nBit) {
// 在Trie中查询与value的xor值最大的答案
// 当前在node这个节点,处理到value的第nBit个bit
// 如果nBit为-1,说明已经处理完毕,node为叶子节点
if (nBit == -1)
return node->value ^ value;
else {
bool bit = ________(5)________;
// 同上
if (node->children[!bit])
return queryTrie(node->children[!bit], value, nBit - 1);
else
return queryTrie(node->children[bit], value, nBit - 1);
}
}

int getAns() {
// 将所有rootXor[i]在Trie中查询最大的xor答案
int ans = 0;
for (int i = 0; i < n; i++)
ans = max(ans, queryTrie(&root, rootXor[i], 30));
return ans;
}

int main() {
while (scanf("%d", &n) == 1) {
readTree();
DFS();
buildTrie();
printf("%d\n", getAns());
}
return 0;
}

注意:这是练习题,不是作业题

全局题号
2975
添加于
2017-09-06
提交次数
13
尝试人数
3
通过人数
3