HDU 1011 Starship Troopers (树形DP)

http://acm.hdu.edu.cn/showproblem.php?pid=1011&PHPSESSID=nrt5ggjfqbhpk9au99mgcvm226

树形DP

题意：给你一颗树，n个顶点，n-1条边，m个花费，从根(编号1)出发，每个顶点有一定花费也有一定收获，并且经过的边不能再次经过，求最大收获是多少？

典型的树形DP，dp[root][k]表示以root为根的子树花费k得到的最大收获，则转移方程为：

dp[root][k]=max(dp[root][k],dp[root][k-i]+dp[t][i])(t为以root的子节点)

注意特判m=0的情况(直接输出0)

  1 // #pragma comment(linker, "/STACK:102400000,102400000")
  2 #include <cstdio>
  3 #include <iostream>
  4 #include <cstring>
  5 #include <string>
  6 #include <cmath>
  7 #include <set>
  8 #include <list>
  9 #include <map>
 10 #include <iterator>
 11 #include <cstdlib>
 12 #include <vector>
 13 #include <queue>
 14 #include <stack>
 15 #include <algorithm>
 16 #include <functional>
 17 using namespace std;
 18 typedef long long LL;
 19 #define ROUND(x) round(x)
 20 #define FLOOR(x) floor(x)
 21 #define CEIL(x) ceil(x)
 22 const int maxn = 110;
 23 const int maxm = 210;
 24 const int inf = 0x3f3f3f3f;
 25 const LL inf64 = 0x3f3f3f3f3f3f3f3fLL;
 26 const double INF = 1e30;
 27 const double eps = 1e-6;
 28 const int P[4] = {0, 0, -1, 1};
 29 const int Q[4] = {1, -1, 0, 0};
 30 const int PP[8] = { -1, -1, -1, 0, 0, 1, 1, 1};
 31 const int QQ[8] = { -1, 0, 1, -1, 1, -1, 0, 1};
 32 
 33 struct Edge
 34 {
 35     int u, v;
 36     int next;
 37 } edge[maxm];
 38 int head[maxn];
 39 int en;
 40 void addsubedge(int u, int v)
 41 {
 42     edge[en].u = u;
 43     edge[en].v = v;
 44     edge[en].next = head[u];
 45     head[u] = en++;
 46 }
 47 void addedge(int u, int v)
 48 {
 49     addsubedge(u, v);
 50     addsubedge(v, u);
 51 }
 52 struct Node
 53 {
 54     int n, p;
 55 } node[maxn];
 56 bool vis[maxn];
 57 int dp[maxn][maxn];
 58 int n, m;
 59 void init()
 60 {
 61     memset(head, -1, sizeof(head));
 62     en = 0;
 63     memset(vis, 0, sizeof(vis));
 64     memset(dp, 0, sizeof(dp));
 65 }
 66 void input()
 67 {
 68     for (int i = 1; i <= n; i++)
 69         scanf("%d%d", &node[i].n, &node[i].p);
 70     for (int i = 0; i < n - 1; i++)
 71     {
 72         int u, v;
 73         scanf("%d%d", &u, &v);
 74         addedge(u, v);
 75     }
 76 }
 77 void dfs(int s)
 78 {
 79     vis[s] = 1;
 80     int num = ceil((double)node[s].n / 20.0);
 81     for (int i = num; i <= m; i++)
 82         dp[s][i] = node[s].p;
 83     for (int i = head[s]; i != -1; i = edge[i].next)
 84     {
 85         int v = edge[i].v;
 86         if (vis[v]) continue;
 87         dfs(v);
 88         for (int j = m; j >= num; j--)
 89         {
 90             for (int k = 1; j + k <= m; k++)
 91             {
 92                 dp[s][j + k] = max(dp[s][j + k], dp[s][j] + dp[v][k]);
 93             }
 94         }
 95     }
 96 }
 97 void debug()
 98 {
 99     //
100 }
101 void solve()
102 {
103     dfs(1);
104 }
105 void output()
106 {
107     printf("%d\n", dp[1][m]);
108 }
109 int main()
110 {
111     // std::ios_base::sync_with_stdio(false);
112 #ifndef ONLINE_JUDGE
113     freopen("in.cpp", "r", stdin);
114 #endif
115 
116     while (~scanf("%d%d", &n, &m))
117     {
118         if (n == -1 && m == -1) return 0;
119         init();
120         input();
121         if (m == 0)
122         {
123             puts("0");
124             continue;
125         }
126         solve();
127         output();
128     }
129     return 0;
130 }
View Code

 

posted @ 2014-01-28 21:13  xysmlx  阅读(197)  评论(0)  编辑  收藏  举报