HDU 1561 The more, The Better (树形DP)

http://acm.hdu.edu.cn/showproblem.php?pid=1561

数组开小了越界，查了好久才查出来。。。GCC太能容忍错误了。。。坑！！！

题意：

给你一棵树，选中某一个节点花费1，得到它的价值，而且必须选中父节点才能选中子节点，问：在花费m下，最大价值为多少？

最基本的树形DP，和前面的几道题一样

dp[s][k]表示选中s节点，总花费k时的最大价值

状态转移：

for (int k = m; k >= 2; k--)
            for (int j = 1; j < k; j++)
                dp[s][k] = max(dp[s][k], dp[s][k - j] + dp[v][j]);
  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 = 210;
 23 const int maxm = 410;
 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 int n, m;
 34 int dp[maxn][maxn];
 35 struct Edge
 36 {
 37     int u, v;
 38     int next;
 39 } edge[maxm];
 40 int en;
 41 int head[maxn];
 42 bool vis[maxn];
 43 int node[maxn];
 44 void addsubedge(int u, int v)
 45 {
 46     edge[en].u = u;
 47     edge[en].v = v;
 48     edge[en].next = head[u];
 49     head[u] = en++;
 50 }
 51 void addedge(int u, int v)
 52 {
 53     addsubedge(u, v);
 54     //addsubedge(v, u);
 55 }
 56 void init()
 57 {
 58     memset(head, -1, sizeof(head));
 59     memset(dp, 0, sizeof(dp));
 60     en = 0;
 61     memset(node, 0, sizeof(node));
 62 }
 63 void input()
 64 {
 65     int u, w;
 66     for (int i = 1; i <= n; i++)
 67     {
 68         scanf("%d%d", &u, &w);
 69         node[i] = w;
 70         addedge(u, i);
 71     }
 72 }
 73 void dfs(int s)
 74 {
 75     //vis[s] = 1;
 76     dp[s][1] = node[s];
 77     // cout << "dfs s: " << s << endl;
 78     //cout<<"dp[s][1]: "<<s<<" "<<dp[s][1]<<endl;
 79     for (int i = head[s]; i != -1; i = edge[i].next)
 80     {
 81         int v = edge[i].v;
 82         //if (!vis[v]) dfs(v);
 83         dfs(v);
 84         for (int k = m; k >= 2; k--)
 85             for (int j = 1; j < k; j++)
 86                 dp[s][k] = max(dp[s][k], dp[s][k - j] + dp[v][j]);
 87     }
 88 }
 89 void debug()
 90 {
 91     for (int i = 0; i <= n; i++) cout << head[i] << " ";
 92     cout << endl;
 93     for (int i = 0; i <= n; i++)
 94     {
 95         for (int j = 0; j <= m; j++) cout << dp[i][j] << '\t';
 96         cout << endl;
 97     }
 98 }
 99 void solve()
100 {
101     node[0] = 0;
102     dfs(0);
103 }
104 void output()
105 {
106     printf("%d\n", dp[0][m]);
107 }
108 int main()
109 {
110     // std::ios_base::sync_with_stdio(false);
111 #ifndef ONLINE_JUDGE
112     freopen("in.cpp", "r", stdin);
113 #endif
114 
115     while (~scanf("%d%d", &n, &m))
116     {
117         if (!n && !m) return 0;
118         m++;
119         init();
120         input();
121         solve();
122         // debug();
123         output();
124     }
125     return 0;
126 }
View Code

 

posted @ 2014-01-29 13:40  xysmlx  阅读(239)  评论(0)  编辑  收藏  举报