poj2763(lca / RMQ + 线段树)

题目链接: http://poj.org/problem?id=2763

 

题意: 第一行输入 n, q, s 分别为树的顶点个数, 询问/修改个数, 初始位置. 接下来 n - 1 行形如 x, y, w 的输入为点 x, y 之间连边且边权为 w.

接下来 q 行输入, 若输入形式为 1 x y 则为将点 x 的权值修改为 y , 若输入形式为 0 x 则询问 s 到 x 的最短距离为多少. 上一组的 x 为下一组的 s.

 

思路: 若去掉修改边权部分, 则为一个 lca 模板题. 对于修改边权, 直接暴力向下修改 dis 数组可能会 tle . 可以将树映射到线段树上, 那样修改时间可降为为 log(n).

具体操作为, dfs 出每点的入点时间戳 l [MAXN] 和出点时间戳 r [MAXN] , 并记录 dfs 顺序 sol [MAXN] .

将 dis [sol [MAXN] ] 放入 sum 数组建线段树.

那么 x 到根结点的距离为: query(l[x], l[x], 1, n, 1) .

将 x 点的权值由 w 修改到 w' 操作为: updata(l[x], r[x], w - w', 1, n, 1) .

 

代码:

  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <math.h>
  4 #include <string.h>
  5 #define lson l, mid, rt << 1
  6 #define rson mid + 1, r, rt << 1 | 1
  7 using namespace std;
  8 
  9 const int MAXN = 3e5 + 10;
 10 struct node{
 11     int v, w, next;
 12     node(){};
 13     node(int V, int W, int NEXT) : v(V), w(W), next(NEXT){};
 14 }edge[MAXN << 1];
 15 
 16 int dp[MAXN << 1][20];
 17 int a[MAXN], b[MAXN], w[MAXN];
 18 int sum[MAXN << 2], add[MAXN << 2];
 19 int l[MAXN], r[MAXN], sol[MAXN], cnt;
 20 int head[MAXN], dis[MAXN], dep[MAXN], ip, indx;
 21 int first[MAXN], ver[MAXN << 1], deep[MAXN << 1];
 22 
 23 void init(void){
 24     memset(head, -1, sizeof(head));
 25     ip = 0;
 26     cnt = 0;
 27     indx = 0;
 28 }
 29 
 30 void addedge(int u, int v, int w){
 31     edge[ip] = node(v, w, head[u]);
 32     head[u] = ip++;
 33 }
 34 
 35 void dfs(int u, int pre, int h){
 36     dep[u] = h;
 37     ver[++indx] = u;
 38     deep[indx] = h;
 39     first[u] = indx;
 40     for(int i = head[u]; i != -1; i = edge[i].next){
 41         int v = edge[i].v;
 42         if(v == pre) continue;
 43         dis[v] = dis[u] + edge[i].w;
 44         dfs(v, u, h + 1);
 45         ver[++indx] = u;
 46         deep[indx] = h;
 47     }
 48 }
 49 
 50 
 51 void ST(int n){
 52     for(int i = 1; i <= n; i++){
 53         dp[i][0] = i;
 54     }
 55     for(int j = 1; (1 << j) <= n; j++){
 56         for(int i = 1; i + (1 << j) - 1 <= n; i++){
 57             int x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j - 1];
 58             dp[i][j] = deep[x] < deep[y] ? x : y;
 59         }
 60     }
 61 }
 62 
 63 int RMQ(int l, int r){
 64     int len = log2(r - l + 1);
 65     int x = dp[l][len], y = dp[r - (1 << len) + 1][len];
 66     return deep[x] < deep[y] ? x : y;
 67 }
 68 
 69 int LCA(int x, int y){
 70     int l = first[x], r = first[y];
 71     if(l > r) swap(l, r);
 72     int pos = RMQ(l, r);
 73     return ver[pos];
 74 }
 75 
 76 void dfs2(int x, int pre){
 77     sol[++cnt] = x;
 78     l[x] = cnt;
 79     for(int i = head[x]; i != -1; i = edge[i].next){
 80         int v = edge[i].v;
 81         if(v != pre) dfs2(v, x);
 82     }
 83     r[x] = cnt;
 84 }
 85 
 86 void push_up(int rt){
 87     sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
 88 }
 89 
 90 void push_down(int rt, int m){
 91     if(add[rt]){
 92         add[rt << 1] += add[rt];
 93         add[rt << 1 | 1] += add[rt];
 94         sum[rt << 1] += (m - (m >> 1)) * add[rt];
 95         sum[rt << 1 | 1] += (m >> 1) * add[rt];
 96         add[rt] = 0;
 97     }
 98 }
 99 
100 void build(int l, int r, int rt){
101     add[rt] = 0;
102     if(l == r){
103         sum[rt] = dis[sol[l]];
104         return;
105     }
106     int mid = (l + r) >> 1;
107     build(lson);
108     build(rson);
109     push_up(rt);
110 }
111 
112 void updata(int L, int R, int key, int l, int r, int rt){
113     if(L <= l && R >= r){
114         sum[rt] += (r - l + 1) * key;
115         add[rt] += key;
116         return;
117     }
118     push_down(rt, r - l + 1);
119     int mid = (l + r) >> 1;
120     if(L <= mid) updata(L, R, key, lson);
121     if(R > mid) updata(L, R, key, rson);
122     push_up(rt);
123 }
124 
125 int query(int L, int R, int l, int r, int rt){
126     if(L <= l && R >= r) return sum[rt];
127     push_down(rt, r - l + 1);
128     int mid = (l + r) >> 1;
129     int ans = 0;
130     if(L <= mid) ans += query(L, R, lson);
131     if(R > mid) ans += query(L, R, rson);
132     return ans;
133 }
134 
135 int main(void){
136     int n, q, s, x, y, z, op;
137     while(~scanf("%d%d%d", &n, &q, &s)){
138         init();
139         for(int i = 1; i < n; i++){
140             scanf("%d%d%d", &x, &y, &z);
141             addedge(x, y, z);
142             addedge(y, x, z);
143             a[i] = x;
144             b[i] = y;
145             w[i] = z;
146         }
147         dis[1] = 0;
148         dfs(1, -1, 1);
149         ST(indx);
150         dfs2(1, -1);
151         build(1, cnt, 1);
152         while(q--){
153             scanf("%d", &op);
154             if(!op){
155                 scanf("%d", &y);
156                 int lca = LCA(s, y);
157                 int sol1 = query(l[s], l[s], 1, cnt, 1);
158                 int sol2 = query(l[y], l[y], 1, cnt, 1);
159                 int sol3 = query(l[lca], l[lca], 1, cnt, 1);
160                 printf("%d\n", sol1 + sol2 - 2 * sol3);
161                 s = y;
162             }else{
163                 scanf("%d%d", &x, &y);
164                 int u = a[x], v = b[x], add = y - w[x];
165                 int cnt2 = dep[u] > dep[v] ? u : v;
166                 updata(l[cnt2], r[cnt2], add, 1, cnt, 1);
167                 w[x] = y;//注意记录修改后的权值
168             }
169         }
170     }
171     return 0;
172 }
View Code

 

posted @ 2017-07-23 15:38  geloutingyu  阅读(368)  评论(0)  编辑  收藏  举报