HDU5293(SummerTrainingDay13-B Tree DP + 树状数组 + dfs序)

最新推荐文章于 2025-09-05 16:34:31 发布

weixin_30347335

最新推荐文章于 2025-09-05 16:34:31 发布

阅读量103

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/Penn000/p/7517896.html

本文介绍了一种解决树链问题的方法，通过深度优先搜索（DFS）确定每个节点的进入和退出时间，使用最近公共祖先（LCA）算法来快速找到任意两节点间的最短路径，并结合树状数组实现高效的区间查询更新，最终求得树上不相交链的最大权重和。

Tree chain problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1798 Accepted Submission(s): 585

Problem Description

Coco has a tree, whose vertices are conveniently labeled by 1,2,…,n.
There are m chain on the tree, Each chain has a certain weight. Coco would like to pick out some chains any two of which do not share common vertices.
Find out the maximum sum of the weight Coco can pick

Input

The input consists of several test cases. The first line of input gives the number of test cases T (T<=10).
For each tests:
First line two positive integers n, m.(1<=n,m<=100000)
The following (n - 1) lines contain 2 integers ai bi denoting an edge between vertices ai and bi (1≤ai,bi≤n),
Next m lines each three numbers u, v and val（1≤u，v≤n，0<val<1000）, represent the two end points and the weight of a tree chain.

Output

For each tests:
A single integer, the maximum number of paths.

Sample Input

1 7 3 1 2 1 3 2 4 2 5 3 6 3 7 2 3 4 4 5 3 6 7 3

Sample Output

Hint

Stack expansion program： #pragma comment(linker, "/STACK:1024000000,1024000000")

Author

FZUACM

Source

2015 Multi-University Training Contest 1

对于每条链u，v，w，我们只在lca(u,v)的顶点上处理它

让dp[i]表示以i为根的子树的最大值，sum[i]表示dp[vi]的和(vi为i的儿子们）

则i点有两种决策，一种是不选以i为lca的链，则dp[i]=sum[i]。

另一种是选一条以i为lca的链，那么有转移方程：dp[i]=sigma(dp[vj])+sigma(sum[kj])+w。(sigma表示累加，vj表示那些不在链上的孩子们，kj表示在链上的孩子们）

为了便于计算，我们处理出dp[i]=sum[i]-sigma(dp[k]-sum[k])+w=sum[i]+sigma(sum[k]-dp[k])+w。

利用dfs序和树状数组可以logn算出sigma(sum[k]-dp[k])。

  1 //2017-09-13
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <vector>
  7 #pragma comment(linker, "/STACK:1024000000,1024000000")
  8 
  9 using namespace std;
 10 
 11 const int N = 210000;
 12 const int LOG_N = 22;
 13 
 14 int head[N], tot;
 15 struct Edge{
 16     int v, next;
 17 }edge[N<<1];
 18 
 19 void add_edge(int u, int v){
 20     edge[tot].v = v;
 21     edge[tot].next = head[u];
 22     head[u] = tot++;
 23 }
 24 
 25 int in[N], out[N], idx, depth[N], father[N][LOG_N];
 26 void dfs(int u, int fa){
 27     in[u] = ++idx;
 28     for(int i = head[u]; i != -1; i = edge[i].next){
 29         int v = edge[i].v;
 30         if(v == fa)continue;
 31         depth[v] = depth[u]+1;
 32         father[v][0]= u;
 33         for(int j = 1; j < LOG_N; j++)
 34           father[v][j] = father[father[v][j-1]][j-1];
 35         dfs(v, u);
 36     }
 37     out[u] = ++idx;
 38 }
 39 
 40 int tree[N];
 41 
 42 int lowbit(int x){
 43     return x&(-x);
 44 }
 45 
 46 void add(int pos, int val){
 47     for(int i = pos; i <= N; i+=lowbit(i))
 48       tree[i] += val;
 49 }
 50 
 51 int query(int l){
 52     int sum = 0;
 53     for(int i = l; i > 0; i-=lowbit(i))
 54       sum += tree[i];
 55     return sum;
 56 }
 57 
 58 int lca(int u, int v){
 59     if(depth[u] < depth[v])
 60       swap(u, v);
 61     for(int i = LOG_N-1; i >= 0; i--){
 62         if(depth[father[u][i]] >= depth[v])
 63           u = father[u][i];
 64     }
 65     if(u == v)return u;
 66     for(int i = LOG_N-1; i >= 0; i--){
 67         if(father[u][i] != father[v][i]){
 68             u = father[u][i];
 69             v = father[v][i];
 70         }
 71     }
 72     return father[u][0];
 73 }
 74 struct Chain{
 75     int u, v, w;
 76 }chain[N];
 77 vector<int> vec[N];
 78 
 79 int dp[N], sum[N];
 80 void solve(int u, int fa){
 81     dp[u] = sum[u] = 0;
 82     for(int i = head[u]; i != -1; i = edge[i].next){
 83         int v = edge[i].v;
 84         if(v == fa)continue;
 85         solve(v, u);
 86         sum[u] += dp[v];
 87     }
 88     dp[u] = sum[u];
 89     for(auto &pos: vec[u]){
 90         int a = chain[pos].u;
 91         int b = chain[pos].v;
 92         int c = chain[pos].w;
 93         dp[u] = max(dp[u], sum[u]+query(in[a])+query(in[b])+c);
 94     }
 95     add(in[u], sum[u]-dp[u]);
 96     add(out[u], dp[u]-sum[u]);
 97 }
 98 
 99 int T, n, m;
100 void init(){
101     tot = 0;
102     idx = 0;
103     depth[1] = 1;
104     for(int i = 1; i <= n; i++)
105       vec[i].clear();
106     memset(head, -1, sizeof(head));
107     memset(dp, 0, sizeof(0));
108     memset(sum, 0, sizeof(0));
109     memset(tree, 0, sizeof(tree));
110 }
111 
112 int main()
113 {
114     freopen("inputB.txt", "r", stdin);
115     scanf("%d", &T);
116     while(T--){
117         scanf("%d%d", &n, &m);
118         init();
119         int u, v;
120         for(int i = 0; i < n-1; i++){
121             scanf("%d%d", &u, &v);
122             add_edge(u, v);
123             add_edge(v, u);
124         }
125         dfs(1, 0);
126         for(int i = 0; i < m; i++){
127             scanf("%d%d%d", &chain[i].u, &chain[i].v, &chain[i].w);
128             vec[lca(chain[i].u, chain[i].v)].push_back(i);
129         }
130         solve(1, 0);
131         printf("%d\n", dp[1]);
132     }
133 
134     return 0;
135 }