Housewife Wind
| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 11419 | Accepted: 3140 |
Description
After their royal wedding, Jiajia and Wind hid away in XX Village, to enjoy their ordinary happy life. People in XX Village lived in beautiful huts. There are some pairs of huts connected by bidirectional roads. We say that huts in the same pair directly connected. XX Village is so special that we can reach any other huts starting from an arbitrary hut. If each road cannot be walked along twice, then the route between every pair is unique.
Since Jiajia earned enough money, Wind became a housewife. Their children loved to go to other kids, then make a simple call to Wind: ‘Mummy, take me home!‘
At different times, the time needed to walk along a road may be different. For example, Wind takes 5 minutes on a road normally, but may take 10 minutes if there is a lovely little dog to play with, or take 3 minutes if there is some unknown strange smell surrounding the road.
Wind loves her children, so she would like to tell her children the exact time she will spend on the roads. Can you help her?
Input
The first line contains three integers n, q, s. There are n huts in XX Village, q messages to process, and Wind is currently in hut s. n < 100001 , q < 100001.
The following n-1 lines each contains three integers a, b and w. That means there is a road directly connecting hut a and b, time required is w. 1<=w<= 10000.
The following q lines each is one of the following two types:
Message A: 0 u
A kid in hut u calls Wind. She should go to hut u from her current position.
Message B: 1 i w
The time required for i-th road is changed to w. Note that the time change will not happen when Wind is on her way. The changed can only happen when Wind is staying somewhere, waiting to take the next kid.
Output
For each message A, print an integer X, the time required to take the next child.
Sample Input
3 3 1 1 2 1 2 3 2 0 2 1 2 3 0 3
Sample Output
1 3
题目链接：POJ 2763
给定一棵树，两种操作，一种是询问当前点到v点的最短距离，并且输出答案之后当前点就变成v点了；第二种是把某一条边的长度变为w。
先考虑只有询问的情况下怎么做，显然是LCA裸题，设dis[u]代表u点到根节点的距离，则有：$dis(u, v) = dis[u]+dis[v]-2*dis[LCA(u,v)]$，然后考虑第二种操作，可以发现当你改变一条边的长度（权值）的时候，会对这条边下的整颗子树造成影响，比如u——v的一条边，且u比v更靠近根节点的话，那么造成的影响是v所在整颗子树的影响，即子树是深度较深的点所连接的。由于是整颗子树，明显可以用DFS序来转换后用树状数组或者线段树来维护这个DFS序列即dis数组，操作就是区间更新，单点查询——改变边长度的时候显然整个子树序列区间长度均改变了$w‘-w$。好像这题一般写法是树链剖分，蒟蒻我不会……就用树状数组好了，树状数组怎么区间更新？左端点处+val，右端点+1处-val即可。
ZZ地WA了一下午发现是把D数组当作深度数组来用了……
代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <numeric>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,sum) memset(arr,sum,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 100010;
struct edge
{
int to, nxt, d;
edge() {}
edge(int _to, int _nxt, int _d): to(_to), nxt(_nxt), d(_d) {}
};
edge E[N << 1];
int head[N], tot;
int ver[N << 1], D[N << 1], F[N], sz, dp[N << 1][19];
int W[N], A[N], B[N];
int L[N], R[N], T[N], ts;
bitset<N>vis;
void init()
{
CLR(head, -1);
tot = 0;
sz = 0;
ts = 0;
CLR(T, 0);
vis.reset();
}
void update(int k, int v)
{
while (k < N)
{
T[k] += v;
k += (k & -k);
}
}
int query(int k)
{
int ret = 0;
while (k)
{
ret += T[k];
k -= (k & -k);
}
return ret;
}
inline void add(int s, int t, int d)
{
E[tot] = edge(t, head[s], d);
head[s] = tot++;
}
void dfs(int u, int d, int sumdis)
{
ver[++sz] = u;
D[sz] = d;
F[u] = sz;
L[u] = ++ts;
update(L[u], sumdis);
update(L[u] + 1, -sumdis);
vis[u] = 1;
for (int i = head[u]; ~i; i = E[i].nxt)
{
int v = E[i].to;
if (!vis[v])
{
dfs(v, d + 1, sumdis + E[i].d);
ver[++sz] = u;
D[sz] = d;
}
}
R[u] = ts;
}
void RMQinit(int l, int r)
{
int i, j;
for (i = l; i <= r; ++i)
dp[i][0] = i;
for (j = 1; l + (1 << j) - 1 <= r; ++j)
{
for (i = l; i + (1 << j) - 1 <= r; ++i)
{
int a = dp[i][j - 1], b = dp[i + (1 << (j - 1))][j - 1];
dp[i][j] = D[a] < D[b] ? a : b;
}
}
}
int LCA(int u, int v)
{
int l = F[u], r = F[v];
if (l > r)
swap(l, r);
int len = r - l + 1;
int k = 0;
while (1 << (k + 1) <= len)
++k;
int a = dp[l][k], b = dp[r - (1 << k) + 1][k];
return D[a] < D[b] ? ver[a] : ver[b];
}
int main(void)
{
int n, q, s, i, a, b, w;
while (~scanf("%d%d%d", &n, &q, &s))
{
init();
for (i = 1; i <= n - 1; ++i)
{
scanf("%d%d%d", &a, &b, &w);
A[i] = a;
B[i] = b;
W[i] = w;
add(a, b, w);
add(b, a, w);
}
dfs(s, 1, 0);
RMQinit(1, sz);
while (q--)
{
int ops;
scanf("%d", &ops);
if (ops == 0)
{
int v;
scanf("%d", &v);
printf("%d\n", query(L[s]) + query(L[v]) - 2 * query(L[LCA(s, v)]));
s = v;
}
else
{
int k, neww;
scanf("%d%d", &k, &neww);
int u = A[k], v = B[k];
if (L[u] > L[v])
swap(u, v);
update(L[v], neww - W[k]);
update(R[v] + 1, -(neww - W[k]));
W[k] = neww;
}
}
}
return 0;
}