Problem Description
有n座城市,由n-1条路相连通,使得任意两座城市之间可达。每条路有过路费,要交过路费才能通过。每条路的过路费经常会更新,现问你,当前情况下,从城市a到城市b最少要花多少过路费。
Input
有多组样例,每组样例第一行输入两个正整数n,m(2 <= n<=50000,1<=m <= 50000),接下来n-1行,每行3个正整数a
b c,(1 <= a,b <= n , a != b , 1 <= c <=
1000000000).数据保证给的路使得任意两座城市互相可达。接下来输入m行,表示m个操作,操作有两种:一. 0 a
b,表示更新第a条路的过路费为b,1 <= a <= n-1 ; 二. 1 a b , 表示询问a到b最少要花多少过路费。
Output
对于每个询问,输出一行,表示最少要花的过路费。
Sample Input
2 3
1 2 1
1 1 2
0 1 2
1 2 1
Sample Output
1
2
思路
以前做树链剖分都是点有取值,这一题是边有权值,我们在处理的时候,对于每一条边的权值,可以把这个权值赋值给比深度比较大的点上。这样就可以把边权改为点权,更容易维护.
这个题令我RE了很久,最后debug成功后发现是int改成long long 造成的re,不知道原因。。
还有一点是w[id[e[i].u]] = e[i].w;
要注意一下
代码
#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <sstream>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#include <list>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e5 + 10;
const int inf = 0x3f3f3f3f;
int first[N], tot, n, m;
int sum[N << 2];
int w[N], cnt, siz[N], dep[N], fa[N], son[N], top[N], id[N];
struct edge
{
int u, v, w, next;
} e[N];
void add_edge(int u, int v, int w)
{
e[tot].u = u, e[tot].v = v, e[tot].w = w;
e[tot].next = first[u];
first[u] = tot++;
}
void init()
{
mem(first, -1);
mem(w, 0);
tot = 0;
cnt = 0;
}
void pushup(int rt)
{
sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void build(int l, int r, int rt)
{
if (l == r)
{
sum[rt] = w[l];
return;
}
int m = (l + r) >> 1;
build(lson);
build(rson);
pushup(rt);
}
void update(int p, int sc, int l, int r, int rt)
{
if (l == r)
{
sum[rt] = sc;
return;
}
int m = (l + r) >> 1;
if (p <= m)
update(p, sc, lson);
else
update(p, sc, rson);
pushup(rt);
}
int query(int L, int R, int l, int r, int rt)
{
if (L <= l && r <= R)
return sum[rt];
int m = (l + r) >> 1;
int ans = 0;
if (L <= m)
ans += query(L, R, lson);
if (R > m)
ans += query(L, R, rson);
pushup(rt);
return ans;
}
void dfs1(int u, int f, int deep)
{
siz[u] = 1;
dep[u] = deep;
fa[u] = f;
son[u] = 0;
int maxson = -1;
for (int i = first[u]; ~i; i = e[i].next)
{
int v = e[i].v;
if (v == f)
continue;
dfs1(v, u, deep + 1);
siz[u] += siz[v];
if (siz[v] > maxson)
{
son[u] = v;
maxson = siz[v];
}
}
}
void dfs2(int u, int topf)
{
top[u] = topf;
id[u] = ++cnt;
if (!son[u])
return;
dfs2(son[u], topf);
for (int i = first[u]; ~i; i = e[i].next)
{
int v = e[i].v;
if (v == fa[u] || v == son[u])
continue;
dfs2(v, v);
}
}
int qsum(int x, int y)
{
int ans = 0;
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]])
swap(x, y);
ans += query(id[top[x]], id[x], 1, n, 1);
x = fa[top[x]];
}
if (dep[x] > dep[y])
swap(x, y);
ans += query(id[son[x]], id[y], 1, n, 1);
return ans;
}
void change(int x, int k)
{
x--;
update(id[e[2 * x].u], k, 1, n, 1); //因为是双向边,所以是e[2*x].u
}
int main()
{
//freopen("in.txt", "r", stdin);
int u, v, c;
while (~scanf("%d%d", &n, &m))
{
init();
for (int i = 1; i <= n - 1; i++)
{
scanf("%d%d%d", &u, &v, &c);
add_edge(u, v, c);
add_edge(v, u, c);
}
dfs1(1, 0, 1);
dfs2(1, 1);
for (int i = 0; i < tot; i += 2)
{
if (dep[e[i].u] < dep[e[i].v])
swap(e[i].u, e[i].v);
w[id[e[i].u]] = e[i].w; //易错点,是w[id[e[i].u]]
}
build(1, n, 1);
while (m--)
{
int op, x, y;
scanf("%d%d%d", &op, &x, &y);
if (op == 0)
change(x, y);
else
printf("%d\n", qsum(x, y));
}
}
return 0;
}