HDU5029--Relief grain（树链剖分）

Problem Description

The soil is cracking up because of the drought and the rabbit kingdom is facing a serious famine. The RRC(Rabbit Red Cross) organizes the distribution of relief grain in the disaster area.

We can regard the kingdom as a tree with n nodes and each node stands for a village. The distribution of the relief grain is divided into m phases. For each phases, the RRC will choose a path of the tree and distribute some relief grain of a certain type for every village located in the path.

There are many types of grains. The RRC wants to figure out which type of grain is distributed the most times in every village.

Input

The input consists of at most 25 test cases.

For each test case, the first line contains two integer n and m indicating the number of villages and the number of phases.

The following n-1 lines describe the tree. Each of the lines contains two integer x and y indicating that there is an edge between the x-th village and the y-th village.
 
The following m lines describe the phases. Each line contains three integer x, y and z indicating that there is a distribution in the path from x-th village to y-th village with grain of type z. (1 <= n <= 100000, 0 <= m <= 100000, 1 <= x <= n, 1 <= y <= n, 1 <= z <= 100000)

The input ends by n = 0 and m = 0.

Output

For each test case, output n integers. The i-th integer denotes the type that is distributed the most times in the i-th village. If there are multiple types which have the same times of distribution, output the minimal one. If there is no relief grain in a village, just output 0.

Sample Input

2 4 1 2 1 1 1 1 2 2 2 2 2 2 2 1 5 3 1 2 3 1 3 4 5 3 2 3 3 1 5 2 3 3 3 0 0

Sample Output

1
2
2
3
3
0
2

 

思路：如果是在区间上的，显然我们可以建立一个棵粮食的线段树，然后每个更新[l,r,z]我们就在l处推入z,r+1处推入-z。最后先扫描一下就行了。

现在问题是在树上的，树链剖分就可以将u,v拆成若干段线段，按树链剖分的遍历顺序扫描一遍就可以了。

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define maxn 400080
#define maxm 800010
#define lson id<<1,l,mid
#define rson id<<1|1,mid+1,r
#define inf 0x3f3f3f3f
int Size[maxn],Son[maxn],dep[maxn],Pos[maxn],fPos[maxn],Top[maxn],fa[maxn],out[maxn];
int first[maxn],nxt[maxm],vv[maxm];
int e,pos;
struct Edge
{
	int u,v,z;
	Edge(){}
	Edge(int uu,int vv,int zz)
	{
		u = uu,v = vv,z = zz;
	}
};
vector  ans;
vector  fuck[maxn];
void init(int n)
{
	e = pos = 0;
	memset(first,-1,sizeof(first));
	memset(Son,-1,sizeof(Son));
	ans.clear();
	for(int i = 1;i <= n+1;i++)
		fuck[i].clear();
}
void addedge(int u,int v)
{
	vv[e] = v;	nxt[e] = first[u];	first[u] = e++;
	vv[e] = u;	nxt[e] = first[v];	first[v] = e++;
}
struct ST
{
	int l,r,key,keynum;
}st[maxn<<2];

void PushUp(int id)
{
	if(st[id<<1|1].key > st[id<<1].key)
		st[id].keynum = st[id<<1|1].keynum;
	else st[id].keynum = st[id<<1].keynum;
	st[id].key = max(st[id<<1].key,st[id<<1|1].key);
}

void buildtree(int id,int l,int r)
{
	st[id].l = l,st[id].r = r;
	st[id].key = 0;
	if(l == r)	
	{
		st[id].keynum = l;
		return;
	}
	int mid = (l+r) >> 1;
	buildtree(lson);
	buildtree(rson);
	PushUp(id);
}
void Update(int id,int pos,int add)
{
	if(st[id].l == pos && st[id].r == pos)
	{
		st[id].key += add;
		return;
	}
	if(st[id<<1].r >= pos)
		Update(id<<1,pos,add);
	else Update(id<<1|1,pos,add);
	PushUp(id);
}

void dfs(int u,int pre)
{
	dep[u] = dep[pre]+1;
	Size[u] = 1;
	for(int i = first[u];i != -1;i = nxt[i])
	{
		int v = vv[i];
		if(v == pre)	continue;
		fa[v] = u;
		dfs(v,u);
		Size[u] += Size[v];
		if(Son[u] == -1 || Size[v] > Size[Son[u]])
			Son[u] = v;
	}
}

void dfs2(int u,int t)
{
	Pos[u] = ++pos;
	fPos[pos] = u;
	Top[u] = t;
	if(Son[u] != -1)
		dfs2(Son[u],t);
	for(int i = first[u];i != -1;i = nxt[i])
	{
		int v = vv[i];
		if(v != fa[u] && v != Son[u])
			dfs2(v,v);
	}
}

void Get_Edge(int u,int v,int z)
{
	int f1 = Top[u],f2 = Top[v];
	while(f1 != f2)
	{
		if(dep[f1] < dep[f2])
		{
			swap(u,v);
			swap(f1,f2);
		}
		ans.push_back(Edge(Pos[f1],Pos[u],z));
		u = fa[f1];	f1 = Top[u];
	}
	if(dep[u] > dep[v])	swap(u,v);
	ans.push_back(Edge(Pos[u],Pos[v],z));
}

void scanf_f(int & a)
{
	a = 0;
	char c;
	while((c = getchar()) < '0' || c > '9');
	a = c - '0';
	while((c = getchar()) >= '0' && c <= '9')
		a = a*10 + c - '0';
}

int main()
{
	//freopen("in.txt","r",stdin);
	int n,m;
	while(scanf("%d%d",&n,&m)==2 && (n|m))
	{
		init(n);
		for(int i = 1;i < n;i++)
		{
			int u,v;
			//scanf("%d%d",&u,&v);
			scanf_f(u);
			scanf_f(v);
			addedge(u,v);
		}
		dep[1] = 0;
		dfs(1,1);
		dfs2(1,1);
		int maxz = 1;
		for(int i = 1;i <= m;i++)
		{
			int x,y,z;
			//scanf("%d%d%d",&x,&y,&z);
			scanf_f(x);
			scanf_f(y);
			scanf_f(z);
			Get_Edge(x,y,z);
			maxz = max(maxz,z);
		}
		int len = ans.size();
		for(int i = 0;i < len;i++)
		{
			int u = ans[i].u,v = ans[i].v,z = ans[i].z;
			fuck[u].push_back(z);
			fuck[v+1].push_back(-z);
		}
		for(int i = 1;i <= n;i++)
			sort(fuck[i].begin(),fuck[i].end());
		buildtree(1,1,maxz);
		for(int i = 1;i <= n;i++)
		{
			int sz = fuck[i].size();
			for(int j = 0;j < sz;j++)
			{
				int k = fuck[i][j];
				if(k < 0)
				{
					Update(1,-k,-1);
				}
				else 
				{
					Update(1,k,1);
				}
			}
			out[fPos[i]] = st[1].keynum;
			if(st[1].key == 0)	out[fPos[i]] = 0;
		}
		for(int i = 1;i <= n;i++)
			printf("%d\n",out[i]);
	}
	return 0;
}