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
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.
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.
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
6
Hint
Stack expansion program： #pragma comment(linker, "/STACK:1024000000,1024000000")Author
FZUACM
Source
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 ", dp[1]); 132 } 133 134 return 0; 135 }