POJ 2763 LCA+BIT

显然这个题是个好题，　
需要修改边利用，前缀和思想动态维护，　每次查询ｕ到ｖ的距离
记下每个点（u）第一次在dfs出现及最后回来的位置，　strart和finish
那么u连向其父亲的边在被修改是影响的只是start【u】和finish【u】之间的范围前缀和恰好可以运用， 还有就是要做一个点到边映射数组， 其实很简单。。哦对了， 相信BIT–树状数组这么6的东西大家都懂吧。。
代码虽然长了， 但思路是很清晰的， 分了两个结构体来做

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

#define N 400010
#define map Map
#define next Next
#define begin Begin
#define C c = getchar()
#define mem(a, b) memset(a, b, sizeof(a))
#define rep(i, s, t) for(int i = s, end = t; i <= end; ++i)
#define erep(i, u) for(int i = begin[u]; i != -1; i = next[i])

bool vis[N];
int eid[N], map[N], dis[N], start[N], finish[N];
int e, cnt, tot, res, st, n, q;
int to[N], next[N], begin[N];
int dep[N], sum[N], dp[N][30], first[N], id[N], dist[N];

void read(int &x) {
    char C; x = 0; while(c<'0' || c>'9') C;
    while(c >= '0' && c <= '9') x = x*10 + c-'0', C;
}

struct BIT {
    int tree[N];
    int lowbit(int x) {
        return x & -x;
    }

    void update(int i, int x) {
        while(i <= n) {
            tree[i] += x;
            i += lowbit(i);
        }
    }

    int sum(int end) {
        int ans = 0;
        while(end > 0) {
            ans += tree[end];
            end -= lowbit(end);
        }
        return ans;
    }
}U;

struct sparse_table {
    void add(int u, int v, int i) {
        eid[++e] = i;
        to[e] = v;
        next[e] = begin[u];
        begin[u] = e;
    }

    void dfs(int u, int depth) {
        int v;
        id[++tot] = u; vis[u] = true;
        first[u] = tot; dep[tot] = depth;
        start[u] = ++cnt;
        erep(i, u)
            if(!vis[v = to[i]]) {
                map[eid[i]] = v;
                dfs(v, depth + 1);
                id[++tot] = u; dep[tot] = depth;
            }
        finish[u] = cnt;
    }

    void ST(int m) {
        rep(i, 1, m) dp[i][0] = i;

        for(int j = 1; (1<<j) <= m; ++j)
            for(int i = 1; i + (1<<j) - 1 <= m; ++i) {
                int a = dp[i][j-1], b = dp[i+(1<<(j-1))][j-1];
                dp[i][j] = dep[a] < dep[b]? a : b;
            }
    }

    int query(int l, int r) {
        int k = 0;
        while(1<<(k+1) <= r - l + 1) k++;
        int x = dp[l][k], y = dp[r - (1<<k) + 1][k];
        return dep[x] < dep[y]? x : y;
    }

    int LCA(int u, int v) {
        int x = first[u], y = first[v];
        if(x > y) swap(x, y);
        return id[query(x, y)];
    }
}T;

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
    freopen("result.out", "w", stdout);
#endif
    mem(begin, -1);
    scanf("%d%d%d", &n, &q, &st);
    rep(i, 1, n-1) {
        int u, v, w;
//      scanf("%d%d%d", &u, &v, &w);
        read(u); read(v); read(w);
        dis[i] = w;
        T.add(u, v, i);
        T.add(v, u, i);
    }

    T.dfs(1, 1);
    T.ST(tot);
    rep(i, 1, n-1) 
        U.update(start[map[i]], dis[i]), U.update(finish[map[i]]+1, -dis[i]);
    rep(i, 1, q) {
        int t, u, w, xx;
//      scanf("%d", &t);
        read(t);
        if(!t) {
//          scanf("%d", &xx);
            read(xx);
            int lca = T.LCA(st, xx);
            printf("%d
", U.sum(start[st]) + U.sum(start[xx]) - (U.sum(start[lca])<<1));
            st = xx;
        }
        else {
    //      scanf("%d%d", &u, &w);
            read(u); read(w);
            U.update(start[map[u]], w - dis[u]);
            U.update(finish[map[u]]+1, dis[u] - w);
            dis[u] = w;
        }
    }
    return 0;
}

好了， LCA的blog就告一段落了， 以后有好题再发