【POJ3237】【树链剖分】Tree

Description

You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:

CHANGE i v	Change the weight of the ith edge to v
NEGATE a b	Negate the weight of every edge on the path from a to b
QUERY a b	Find the maximum weight of edges on the path from a to b

Input

The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.

Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.

Output

For each “QUERY” instruction, output the result on a separate line.

Sample Input

1



3

1 2 1

2 3 2

QUERY 1 2

CHANGE 1 3

QUERY 1 2

DONE

Sample Output

1

3

Source

POJ Monthly--2007.06.03, Lei, Tao

【分析】

好久没写了。

写一道练练手。

  1 /*

  2 宋代苏轼

  3 《南乡子·重九涵辉楼呈徐君猷》

  4 霜降水痕收。浅碧鳞鳞露远洲。酒力渐消风力软，飕飕。破帽多情却恋头。

  5 佳节若为酬。但把清尊断送秋。万事到头都是梦，休休。明日黄花蝶也愁。

  6 */

  7 #include <iostream>

  8 #include <cstdio>

  9 #include <algorithm>

 10 #include <cstring>

 11 #include <vector>

 12 #include <utility>

 13 #include <iomanip>

 14 #include <string>

 15 #include <cmath>

 16 #include <queue>

 17 #include <assert.h>

 18 #include <map>

 19 #include <ctime>

 20 #include <cstdlib>

 21 #include <stack>

 22 #define LOCAL

 23 const int INF = 0x3f3f3f3f;

 24 const int maxn=100000 + 10;

 25 const int maxnode = 100000;

 26 const int maxm=100000 * 2 + 10;

 27 using namespace std;

 28 struct Node{//权值线段树 

 29        int l, r;

 30        int Max, Min;

 31        bool neg;//取反标记 

 32 }tree[maxn * 4];

 33 struct Edge{

 34        int u, v, w;

 35 }edges[maxn];//输入的边 

 36 

 37 int n, fa[maxn], size[maxn];

 38 int son[maxn], dep[maxn], top[maxn];

 39 int pos[maxn], Time;

 40 int M, head[maxn], next[maxm], to[maxm], w[maxm];

 41 

 42 //第一次dfs 

 43 void dfs_1(int u){

 44      size[u] = 1;

 45      son[u] = 0;

 46      for (int i = head[u]; i != -1; i = next[i]){

 47          int v = to[i];

 48          if (v == fa[u]) continue;

 49          dep[v] = dep[u] + 1;

 50          fa[v] = u;

 51          dfs_1(v);

 52          size[u] += size[v];

 53          if (size[v] > size[son[u]]) son[u] = v;

 54      }

 55      return;

 56 }

 57 void dfs_2(int u, int top_node){

 58      top[u] = top_node;

 59      pos[u] = ++Time;//给他和他的父亲的边在线段树中的位置 

 60      //重边

 61      if (son[u]) dfs_2(son[u], top_node); 

 62      //轻边 

 63      for (int i = head[u]; i != -1; i = next[i]){

 64          int v = to[i];

 65          if (v == fa[u] || v == son[u]) continue;

 66          dfs_2(v, v);

 67      }

 68 }

 69 //建树 

 70 void build(int x, int l, int r){

 71      tree[x].l = l;tree[x].r = r;

 72      tree[x].Max = -INF;

 73      tree[x].Min = INF;

 74      tree[x].neg = 0;

 75      if (l == r) return;

 76      

 77      int mid = (l + r) >> 1;

 78      build(x << 1, l, mid);

 79      build((x << 1) | 1, mid + 1, r);

 80 }

 81 void update(int x){

 82      tree[x].Max = max(tree[x << 1].Max, tree[(x << 1) | 1].Max);

 83      tree[x].Min = min(tree[x << 1].Min, tree[(x << 1) | 1].Min);

 84      return;

 85 }

 86 //标记下传 

 87 void pushdown(int x){

 88      if (tree[x].l == tree[x].r) return;

 89      

 90      if (tree[x].neg){

 91         int l = (x << 1), r = l | 1;

 92         tree[l].neg ^= 1;

 93         tree[l].Min *= -1;

 94         tree[l].Max *= -1;

 95         swap(tree[l].Min, tree[l].Max);

 96         

 97         tree[r].neg ^= 1;

 98         tree[r].Min *= -1;

 99         tree[r].Max *= -1;

100         swap(tree[r].Min, tree[r].Max);

101         

102         tree[x].neg = 0;

103      }

104 }

105 //在线段树中修改l,r为val 

106 void change2(int x, int l, int r, int val){

107      pushdown(x);

108      if (tree[x].l == l && tree[x].r == r){

109         if (val == INF){//取反操作,注意已经pushdown过了 

110            tree[x].neg = 1;

111            tree[x].Min *= -1;

112            tree[x].Max *= -1;

113            swap(tree[x].Min, tree[x].Max);

114         } else tree[x].Min = tree[x].Max = val;//更新val 

115         return;

116      }

117      

118      int mid = (tree[x].l + tree[x].r) >> 1;

119      if (r <= mid) change2(x << 1, l, r, val);

120      else if (l > mid) change2((x << 1) | 1, l, r, val);

121      else{

122           change2(x << 1, l, mid, val);

123           change2((x << 1) | 1, mid + 1, r, val);

124      }

125      update(x);

126 }

127 int query2(int x, int l, int r){

128      pushdown(x);

129      if (tree[x].l == l && tree[x].r == r) return tree[x].Max;

130      

131      int mid = (tree[x].l + tree[x].r) >> 1; 

132      if (r <= mid) return query2(x << 1, l, r);

133      else if (l > mid) return query2((x << 1) | 1, l, r);

134      else return max(query2((x << 1), l, mid), query2((x << 1) | 1, mid + 1, r));

135 }

136 //树链剖分部分 

137 void change(int x, int y, int v){

138      while (top[x] != top[y]){

139            //总是矮的往上爬.. 

140            //保证dep[top[x]] >= dep[top[y]] 

141            if (dep[top[x]] < dep[top[y]]) swap(x, y);

142            

143            change2(1, pos[top[x]], pos[x], v);

144            x = fa[top[x]];

145      }

146      

147      if (x == y) return;

148      if (dep[x] > dep[y]) swap(x, y);

149      change2(1, pos[son[x]], pos[y], v);

150 }

151 int query(int x, int y){

152     int Ans = -INF;

153     while (top[x] != top[y]){

154           if (dep[top[x]] < dep[top[y]]) swap(x, y);

155           

156           Ans = max(Ans, query2(1, pos[top[x]], pos[x]));

157           x = fa[top[x]];

158     }

159     if (x == y) return Ans;

160     if (dep[x] > dep[y]) swap(x, y);

161     Ans = max(Ans, query2(1, pos[son[x]], pos[y]));

162     return Ans == -INF ? 0 : Ans;

163 }

164 void work(){

165     while(1){

166         char str[20];

167         scanf("%s", str);

168         if(str[0] == 'C'){

169             int x, v;

170             scanf("%d%d", &x, &v);

171             change2(1, pos[edges[x].v], pos[edges[x].v], v);

172         }else if(str[0] == 'N'){

173             int l, r;

174             scanf("%d%d", &l, &r);

175             change(l, r, INF);

176         }else if(str[0] == 'Q'){//询问两点之间最大值 

177             int l, r;

178             scanf("%d%d", &l, &r);

179             printf("%d\n", query(l, r));

180         }else break;

181     }

182 }

183 //加边 

184 void addEdge(int u, int v, int c){

185      to[M] = v;

186      w[M] = c;

187      next[M] = head[u];

188      head[u] = M++;

189 }

190 void init(){

191      memset(head, -1, sizeof(head));//邻接表初始化

192      memset(dep, 0, sizeof(dep));

193      M = Time = 0;//总边数和时间 

194      fa[1] = size[0] = 0;

195      //加边 

196      scanf("%d", &n);

197      for (int i = 1; i < n; i++){

198          scanf("%d%d%d", &edges[i].u, &edges[i].v, &edges[i].w);

199          addEdge(edges[i].u, edges[i].v, edges[i].w);

200          addEdge(edges[i].v, edges[i].u, edges[i].w);

201      }

202      dfs_1(1);

203      dfs_2(1, 1);

204      build(1, 1, Time);

205      for (int i = 1; i < n; i++){

206          int x = edges[i].u, y = edges[i].v;

207          //判断父亲 

208          if (dep[x] > dep[y]) swap(edges[i].u, edges[i].v);

209          //u一定是父亲 

210          change2(1, pos[edges[i].v], pos[edges[i].v], edges[i].w);

211     }

212 }

213 

214 int main(){

215     int T;

216     

217     scanf("%d", &T);

218     while (T--){

219           init();

220           work();

221     }

222     //printf("%d\n", INF); 

223     return 0;

224 }

View Code