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\n", 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