题目来源：POJ 2763 Housewife Wind

题意：给你一棵树 2种操作0 x 求当前点到x的最短路 然后当前的位置为x; 1 i x 将第i条边的权值置为x

思路：树上两点u, v距离为d[u]+d[v]-2*d[LCA(u,v)] 现在d数组是变化的 对应每一条边的变化 他修改的是一个区间 用时间戳处理每个点管辖的区域 然后用线段树修改 线段树的叶子节点村的是根到每一个点的距离 求最近公共祖先没差别 只是堕落用线段树维护d数组

各种错误 4个小时 伤不起

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 200010;
struct edge
{
	int u, v, w, next;
}edges[maxn*2], e[maxn];

int E[maxn*2], H[maxn*2], I[maxn*2], L[maxn], R[maxn];
int dp[maxn*2][40];
int cnt, clock, dfn;
int first[maxn];
int a[maxn<<2];
int b[maxn];
int add[maxn<<2];
int degree[maxn];
int vis[maxn];
void AddEdge(int u, int v, int w)
{
	edges[cnt].u = u;
	edges[cnt].v = v;
	edges[cnt].w = w;
	edges[cnt].next = first[u];
	first[u] = cnt++;
	edges[cnt].u = v;
	edges[cnt].v = u;
	edges[cnt].w = w;
	edges[cnt].next = first[v];
	first[v] = cnt++;	
}
void dfs(int u, int fa, int dep)
{
	E[++clock] = u;
	H[clock] = dep;
	I[u] = clock;
	L[u] = ++dfn;
	b[dfn] = u;
	for(int i = first[u]; i != -1; i = edges[i].next)
	{
		int v = edges[i].v;
		if(v == fa)
			continue;
		if(vis[v])
			continue;
		vis[v] = true;
		dfs(v, u, dep+1);
		E[++clock] = u;
		H[clock] = dep;
	}
	R[u] = dfn;
}

void RMQ_init(int n)
{
	for(int i = 1; i <= n; i++)
		dp[i][0] = i;
	for(int j = 1; (1<<j) <= n; j++)
		for(int i = 1; i+(1<<j)-1 <= n; i++)
		{
			if(H[dp[i][j-1]] < H[dp[i+(1<<(j-1))][j-1]])
				dp[i][j] = dp[i][j-1];
			else
				dp[i][j] = dp[i+(1<<(j-1))][j-1];
		}
}
int RMQ(int l, int r)
{
	l = I[l], r = I[r];
	if(l > r)
		swap(l, r);
	int len = r-l+1, k = 0;
	while((1<<k) <= len)
		k++;
	k--;
	if(H[dp[l][k]] < H[dp[r-(1<<k)+1][k]])
		return E[dp[l][k]];
	else
		return E[dp[r-(1<<k)+1][k]];
}
void pushdown(int rt, int l, int r)
{
	int k = (r-l+1);
	if(add[rt])
	{
		a[rt<<1] += add[rt]*(k-(k>>1));
		a[rt<<1|1] += add[rt]*(k>>1);
		add[rt<<1] += add[rt];
		add[rt<<1|1] += add[rt];
		add[rt] = 0;
	}
}

void build(int l, int r, int rt)
{
	a[rt] = 0;
	add[rt] = 0;
	if(l == r)
		return;
	int m = (l + r) >> 1;
	build(l, m, rt<<1);
	build(m+1, r, rt<<1|1);
}

void update(int x, int y, int l, int r, int rt, int num)
{
	if(l == x && r == y)
	{
		a[rt] += (r-l+1)*num;
		add[rt] += num;
		return;
	}
	pushdown(rt, l, r);
	int m = (l + r) >> 1;
	if(y <= m)
		update(x, y, l, m, rt<<1, num);
	else if(x > m)
		update(x, y, m+1, r, rt<<1|1, num);
	else
	{
		update(x, m, l, m, rt<<1, num);
		update(m+1, y, m+1, r, rt<<1|1, num);
	}
	a[rt] = a[rt<<1] + a[rt<<1|1];
}


int query(int x, int l, int r, int rt)
{
	if(l == r)
	{
		return a[rt];
	}
	pushdown(rt, l, r);
	int m = (l + r) >> 1;
	int ans = 0;
	if(x <= m)
		ans = query(x, l, m, rt<<1);
	else
		ans = query(x, m+1, r, rt<<1|1);
	a[rt] = a[rt<<1] + a[rt<<1|1];
	return ans;
}
int main()
{
	int cas = 1;
	int T;
	//scanf("%d", &T);
	int s, to, root, n, q;
	while(scanf("%d %d %d", &n, &q, &s) != EOF)
	{
		memset(vis, 0, sizeof(vis));
		memset(first, -1, sizeof(first));
		memset(degree, 0, sizeof(degree));
		clock = cnt = dfn = 0;
		
		build(1, n, 1);
		//for(int i = 1; i <= n; i++)
		//	scanf("%d", &b[i]);
		for(int i = 1; i < n; i++)
		{
			int u, v, w;
			scanf("%d %d %d", &u, &v, &w);
			e[i].u = u;
			e[i].v = v;
			e[i].w = w;
			AddEdge(u, v, 0);
			degree[v]++;
		}
		
		for(int i = 1; i <= n; i++)
			if(!degree[i])
			{
				vis[i] = true;
				dfs(i, -1, 0);
				root = i;
				break;
			}
		RMQ_init(2*n-1);
		//puts("1");
		for(int i = 1; i < n; i++)
		{
			int u = e[i].u;
			int v = e[i].v;
			int w = e[i].w;
			//printf("***%d %d\n", L[v], R[v]);
			if(L[u] < L[v])
				update(L[v], R[v], 1, n, 1, w);
			else
				update(L[u], R[u], 1, n, 1, w);
		}
		
		while(q--)
		{
			int x;
			scanf("%d", &x);
			if(!x)
			{
				scanf("%d", &to);
				int d1 = query(L[s], 1, n, 1);
				int d2 = query(L[to], 1, n, 1);
				int lca = RMQ(s, to);
				int d3 = query(L[lca], 1, n, 1);
				//printf("***%d %d %d\n", d1, d2, d3);
				printf("%d\n", d1+d2-2*d3);
				//printf("%d\n", dfn);
				s = to;
			}
			else
			{
				int i, w;
				scanf("%d %d", &i, &w);
				int x = w - e[i].w;
				e[i].w = w;
				int v = e[i].v;
				int u = e[i].u;
				if(L[u] < L[v])
					update(L[v], R[v], 1, n, 1, x);
				else
					update(L[u], R[u], 1, n, 1, x);
			}
		}
	}
	return 0;
}