BZOJ 2819 Nim 树链剖分

标签：bzoj   bzoj2819   树链剖分   nim游戏   

题目大意：两个小人在树上玩Nim游戏，问有没有必胜策略。

思路：尼姆游戏：如果所有石子的异或值为0就是必败局面。异或有如下性质：x ^ y ^ z = x ^ (y ^ z),所以就可以进行树链剖分了。题目中还好心提醒有30%的点会使dfs爆栈。。第一次写不用dfs的树链剖分，扒的别人的代码，有些丑陋。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 500010
#define LEFT (pos << 1)
#define RIGHT (pos << 1|1)
using namespace std;
const char yes[] = "Yes";
const char no[] = "No";

int points,asks;
int src[MAX],line[MAX];

int head[MAX],total;
int next[MAX << 1],aim[MAX << 1];

int q[MAX];
int deep[MAX],son[MAX],father[MAX],size[MAX];
int top[MAX],pos[MAX],cnt;

int tree[MAX << 2];

char s[10];

inline void Add(int x,int y);
void BFS();

void BuildTree(int l,int r,int _pos);
bool Ask(int x,int y);
int Ask(int l,int r,int x,int y,int pos);
void Modify(int l,int r,int x,int pos,int c);

int main()
{
	cin >> points;
	for(int i = 1;i <= points; ++i)
		scanf("%d",&src[i]);
	for(int x,y,i = 1;i < points; ++i) {
		scanf("%d%d",&x,&y);
		Add(x,y),Add(y,x);
	}
	BFS();
	for(int i = 1;i <= points; ++i)
		line[pos[i]] = src[i];
	BuildTree(1,cnt,1);
	cin >> asks;
	for(int x,y,i = 1;i <= asks; ++i) {
		scanf("%s%d%d",s,&x,&y);
		if(s[0] == 'Q')
			puts(Ask(x,y) ? yes:no);
		else	Modify(1,points,pos[x],1,y);
	}
	return 0;
}

inline void Add(int x,int y)
{
	next[++total] = head[x];
	aim[total] = y;
	head[x] = total;
}

void BFS()
{
	int r = 0,h = 0;
	q[++r] = 1;
	while(r != h) {
		int x = q[++h];
		deep[x] = deep[father[x]] + 1;
		size[x] = 1;
		for(int i = head[x];i;i = next[i]) {
			if(aim[i] == father[x])	continue;
			father[aim[i]] = x;
			q[++r] = aim[i]; 
		}
	}
	for(int i = points;i; --i) {
		int x = q[i];
		size[father[x]] += size[x];
		if(size[x] > size[son[father[x]]])
			son[father[x]] = x;
	}
	for(int i = 1;i <= points; ++i) {
		int x = q[i];
		if(son[father[x]] == x)
			top[x] = top[father[x]];
		else {
			top[x] = x;
			for(int j = x;j;j = son[j])
				pos[j] = ++cnt;
		}
	}
}

void BuildTree(int l,int r,int _pos)
{
	if(l == r) {
		tree[_pos] = line[l];
		return ;
	}
	int mid = (l + r) >> 1;
	BuildTree(l,mid,_pos << 1);
	BuildTree(mid + 1,r,_pos << 1|1);
	tree[_pos] = tree[_pos << 1] ^ tree[_pos << 1|1];
}

bool Ask(int x,int y)
{
	int re = 0;
	while(top[x] != top[y]) {
		if(deep[top[x]] < deep[top[y]])
			swap(x,y);
		re ^= Ask(1,points,pos[top[x]],pos[x],1);
		x = father[top[x]];
	}
	if(deep[x] < deep[y])	swap(x,y);
	re ^= Ask(1,points,pos[y],pos[x],1);
	return re ? true:false;
}

int Ask(int l,int r,int x,int y,int pos)
{
	if(l == x && y == r)	return tree[pos];
	int mid = (l + r) >> 1;
	if(y <= mid)	return Ask(l,mid,x,y,LEFT);
	if(x > mid)		return Ask(mid + 1,r,x,y,RIGHT);
	int left = Ask(l,mid,x,mid,LEFT);
	int right = Ask(mid + 1,r,mid + 1,y,RIGHT);
	return left ^ right;
}

void Modify(int l,int r,int x,int pos,int c)
{
	if(l == r) {
		tree[pos] = c;
		return ;
	}
	int mid = (l + r) >> 1;
	if(x <= mid)	Modify(l,mid,x,LEFT,c);
	else	Modify(mid + 1,r,x,RIGHT,c);
	tree[pos] = tree[LEFT] ^ tree[RIGHT]; 
}

BZOJ 2819 Nim 树链剖分

标签：bzoj   bzoj2819   树链剖分   nim游戏   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/39636723