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题意:两种操作,问u到v的距离,并且u走到了v;把第i条边距离改成w
分析:根据DFS访问顺序,将树处理成链状的,那么回边处理成负权值,那么LCA加上BIT能够知道u到v的距离,BIT存储每条边的信息,这样第二种操作也能用BIT快速解决
利用RMQ的写法不知哪里写挫了,改用倍增法
/************************************************ * Author :Running_Time * Created Time :2015/10/6 星期二 11:45:03 * File Name :POJ_2763.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int D = 20; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-8; struct Edge { int v, w, id, nex; }edge[N<<1]; struct BIT { int c[N<<1], NN; void init(int n) { NN = n * 2; memset (c, 0, sizeof (c)); } void updata(int i, int x) { while (i <= NN) { c[i] += x; i += i & (-i); } } int query(int i) { int ret = 0; while (i) { ret += c[i]; i -= i & (-i); } return ret; } }bit; int head[N]; int dep[N], rt[D][N], id[N], in[N], out[N]; int cost[N]; int e, tim; void init(void) { memset (head, -1, sizeof (head)); e = 0; } void add_edge(int u, int v, int w, int id) { edge[e] = (Edge) {v, w, id, head[u]}; head[u] = e++; } void DFS(int u, int fa, int d) { dep[u] = d; rt[0][u] = fa; for (int i=head[u]; ~i; i=edge[i].nex) { Edge &e = edge[i]; if (e.v == fa) continue; in[e.id] = id[e.v] = ++tim; DFS (e.v, u, d + 1); out[e.id] = ++tim; } } int LCA(int u, int v) { if (dep[u] < dep[v]) { swap (u, v); } for (int i=0; i<D; ++i) { if ((dep[u] - dep[v]) >> i & 1) { u = rt[i][u]; } } if (u == v) return u; for (int i=D-1; i>=0; --i) { if (rt[i][u] != rt[i][v]) { u = rt[i][u]; v = rt[i][v]; } } return rt[0][u]; } int main(void) { int n, q, s; while (scanf ("%d%d%d", &n, &q, &s) == 3) { init (); for (int u, v, w, i=1; i<n; ++i) { scanf ("%d%d%d", &u, &v, &cost[i]); add_edge (u, v, cost[i], i); add_edge (v, u, cost[i], i); } tim = 0; DFS (1, -1, 0); for (int i=1; i<D; ++i) { for (int j=1; j<=n; ++j) { rt[i][j] = rt[i-1][j] == -1 ? -1 : rt[i-1][rt[i-1][j]]; } } bit.init (n); for (int i=1; i<n; ++i) { bit.updata (in[i], cost[i]); //入边序号 bit.updata (out[i], -cost[i]); //回边序号 } int u = s; for (int op, i=1; i<=q; ++i) { scanf ("%d", &op); if (op == 0) { int v; scanf ("%d", &v); //入点序号和回点序号 printf ("%d\n", bit.query (id[u]) + bit.query (id[v]) - bit.query (id[LCA (u, v)]) * 2); u = v; } else { int p, w; scanf ("%d%d", &p, &w); bit.updata (in[p], w - cost[p]); bit.updata (out[p], cost[p] - w); cost[p] = w; } } } return 0; }
LCA+树状数组 POJ 2763 Housewife Wind
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原文地址:http://www.cnblogs.com/Running-Time/p/4857340.html