HDU 3586【树形DP+二分】

题目：Information Disturbing

题意：

给出一棵树，和每条边的cost值，设置一个界限，要求切断所有叶子结点，切断的边的cost不能超过界限值，并且切断的边的总和不能超过m,求出这个最小界限值。

解题思路:

很明显的一道树型DP，DP[v],表示切断根结点的v的子树的最小花费，那么这个界限怎么处理呢，由于界限值范围要比总和m小得多，才1－1000，可以用二分枚举，这样在选择切边时可以根据这个枚举值进行判断处理。每个结点v，设其儿子结点为x1,x2,x3...则DP[v] = Min{DP[x1],Node[x1].cost} + Min{DP[x2], Node[x2].cost} + ...，当然，要判断Node[x].cost是否小于枚举的界限，Node[x].cost表示边（v,x）的cost值。

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <string>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <vector>
 7 #include <map>
 8 
 9 using namespace std;
10 
11 const int MAX = 1000 + 10;
12 const int INF = 1001;
13 const int MAXM = 1000000 + 10;
14 struct TTree
15 {
16     int node, cost, next;
17 }Tree[MAX * 4];
18 
19 int Pos[MAX];
20 int EdgeNumber;
21 int DP[MAX];
22 void init(int n)
23 {
24     EdgeNumber = n;
25     for(int i = 1; i <= n; ++i)
26     {
27         Pos[i] = i;
28         Tree[i].next = 0;
29     }
30 }
31 void addEdge(int from, int to, int cost)
32 {
33     ++EdgeNumber;
34     Tree[Pos[from]].next = EdgeNumber;
35     Tree[Pos[from] = EdgeNumber].node = to; 
36     Tree[EdgeNumber].cost = cost;
37     Tree[EdgeNumber].next = 0;
38 }
39 
40 void dfs(int father, int node, int upper)
41 {
42     int sum = 0;
43     for(int ix = Tree[node].next; ix != 0; ix = Tree[ix].next)
44     {
45         if(Tree[ix].node != father)
46         {
47             dfs(node, Tree[ix].node, upper);
48             if(Tree[ix].cost <= upper && Tree[ix].cost < DP[Tree[ix].node])
49             {
50                 sum += Tree[ix].cost;
51             }
52             else
53                 sum += DP[Tree[ix].node];
54         }
55     }
56     if(sum == 0)
57         sum = MAXM;
58     DP[node] = sum;
59 }
60 
61 int solve(int maxSum)
62 {
63     int left = 1, right = INF;
64     int mid;
65     while(left < right)
66     {
67         mid = (left + right) >> 1;
68         dfs(-1, 1, mid);
69         if(DP[1] <= maxSum)
70             right = mid;
71         else
72             left = mid + 1;
73     }
74     return left;
75 }
76 int main()
77 {
78     freopen("in.txt","r",stdin);
79     int n, m;
80     int x, y, c;
81     while(scanf("%d%d", &n, &m) == 2 && (n || m))
82     {
83         init(n);
84         for(int i = 1; i <= n - 1; ++i)
85         {
86             scanf("%d%d%d", &x, &y, &c);
87             addEdge(x, y, c);
88             addEdge(y, x, c);
89         }
90         int ans = solve(m);
91         if(ans >= INF)
92             printf("-1\n");
93         else
94             printf("%d\n", ans);
95     }
96     return 0;
97 }