BZOJ 2783 JLOI 2012 树 倍增+二分

标签：bzoj   jloi2012   倍增   二分   树   

题目大意：给出一棵树和一个整数s，问在树上有几条这样路径，保证路径上的点权和＝＝s，点的深度递增。输出这个数量。

思路：利用倍增的思想，我们能在O(logn)的时间内求出一个点到他的第n个爸爸之间所有点的点权之和。由于点权只能是正的，满足二分性质。然后对于每一个点二分，看看有没有路径的权值和是S，统计答案，输出。

CODE：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 100010
using namespace std;

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

int father[MAX][20],length[MAX][20];

inline void Add(int x,int y);
void DFS(int x,int last);
void SparseTable();

inline bool Judge(int x);
inline int GetLength(int x,int deep);

int main()
{
	cin >> points >> s;
	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);
	}
	DFS(1,0);
	SparseTable();
	int ans = 0;
	for(int i = 1;i <= points; ++i)
		ans += Judge(i);
	cout << ans << endl;
	return 0;
}

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

void DFS(int x,int last)
{
	father[x][0] = last;
	length[x][0] = src[last];
	for(int i = head[x];i;i = next[i]) {
		if(aim[i] == last)	continue;
		DFS(aim[i],x);
	}
}

void SparseTable()
{
	for(int j = 1;j <= 19; ++j)
		for(int i = 1;i <= points; ++i) {
			father[i][j] = father[father[i][j - 1]][j - 1];
			length[i][j] = length[i][j - 1] + length[father[i][j - 1]][j - 1];
		}
}

inline bool Judge(int x)
{
	int l = 0,r = 100000;
	while(l <= r) {
		int mid = (l + r) >> 1;
		int length = GetLength(x,mid) + src[x];
		if(length < s)	l = mid + 1;
		else if(length > s)	r = mid - 1;
		else	return true;
	}
	return false;
}

inline int GetLength(int x,int deep)
{
	int re = 0;
	for(int i = 19; ~i; --i)
		if(deep - (1 << i) >= 0) {
			deep -= 1 << i;
			re += length[x][i];
			x = father[x][i];
		}
	return re;
}

BZOJ 2783 JLOI 2012 树 倍增+二分

标签：bzoj   jloi2012   倍增   二分   树   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/40106817