Game

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 336    Accepted Submission(s): 92

Problem Description
　　Nowadays, there are more and more challenge game on TV such as 'Girls, Rush Ahead'. Now, you participate int a game like this. There are N rooms. The connection of rooms is like a tree. In other words, you can go to any other room by one and only one way. There is a gift prepared for you in Every room, and if you go the room, you can get this gift. However, there is also a trap in some rooms. After you get the gift, you may be trapped. After you go out a room, you can not go back to it any more. You can choose to start at any room ,and when you have no room to go or have been trapped for C times, game overs. Now you would like to know what is the maximum total value of gifts you can get.
 
Input
　　The first line contains an integer T, indicating the number of testcases. 　　For each testcase, the first line contains one integer N(2 <= N <= 50000), the number rooms, and another integer C(1 <= C <= 3), the number of chances to be trapped. Each of the next N lines contains two integers, which are the value of gift in the room and whether have trap in this rooom. Rooms are numbered from 0 to N-1. Each of the next N-1 lines contains two integer A and B(0 <= A,B <= N-1), representing that room A and room B is connected. 　　All gifts' value are bigger than 0.
 
Output
　　For each testcase, output the maximum total value of gifts you can get.
 
Sample Input
2 3 1 23 0 12 0 123 1 0 2 2 1 3 2 23 0 12 0 123 1 0 2 2 1
 
Sample Output
146 158
 
Source
 
Recommend
zhuyuanchen520
 

思路：dp[u][trap][dir]最优值，起始节点，已经过的trap数，是入点还是出点。0 out 1 in

ans=max(dp[u][trapu][0]+dp[v][trapv][1],dp[u][trapu][1]+dp[v][trapv][0],dp[u][trapu][0]+dp[v][trapv][0])

注意trap==C

hdu 4616 Game (树形DP，4级)_javahdu 4616 Game (树形DP，4级)_java_02
 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #define FOR(i,n) for(int i=0;i<n;++i)
 5 #define clr(f,z) memset(f,z,sizeof(f))
 6 using namespace std;
 7 const int mm=5e4+9;
 8 class Edge
 9 {
10   public:int v,next;
11 }e[mm*2];
12 class node
13 {
14   public:int trap,val;
15 }f[mm];
16 int head[mm],edge;
17 bool vis[mm];
18 int cas,n,C,ans,dp[mm][5][2];
19 void data()
20 {
21   clr(head,-1);edge=0;
22 }
23 void add(int u,int v)
24 {
25   e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
26 }
27 void DP(int u)
28 {
29   int v,trap;vis[u]=1;
30   trap=f[u].trap;
31   dp[u][trap][1]=dp[u][trap][0]=f[u].val;
32   for(int i=head[u];~i;i=e[i].next)
33   {
34     v=e[i].v;
35     if(vis[v])continue;
36     DP(v);
37     for(int j=0;j<=C;++j)
38       for(int k=0;k+j<=C;++k)
39     {
40       if(j!=C)///0 out if(j==C) stop at u can't reatch v
41       {
42         ans=max(ans,dp[u][j][0]+dp[v][k][1]);
43       }
44       if(k!=C)///1 in if(k==C) stop at v can't reatch u
45       {
46         ans=max(ans,dp[u][j][1]+dp[v][k][0]);
47       }
48       if(j+k<C)
49       {
50         ans=max(ans,dp[u][j][0]+dp[v][k][0]);
51       }
52     }
53     trap=f[u].trap;
54     for(int j=0;j<=C;++j)
55       {
56         dp[u][j+trap][0]=max(dp[u][j+trap][0],dp[v][j][0]+f[u].val);
57       }
58     for(int j=1;j<=C;++j)///1 in if(j==0) it's impossble to reatch u
59     {
60       dp[u][j+trap][1]=max(dp[u][j+trap][1],dp[v][j][1]+f[u].val);
61     }
62   }
63 }
64 int main()
65 { int a,b;
66   while(~scanf("%d",&cas))
67   {
68     while(cas--)
69     { data();
70       scanf("%d%d",&n,&C);
71       FOR(i,n)
72       scanf("%d%d",&f[i].val,&f[i].trap);
73       FOR(i,n-1)
74       {
75         scanf("%d%d",&a,&b);add(a,b);add(b,a);
76       }
77       clr(vis,0);clr(dp,0);
78       ans=0;
79       DP(0);
80       printf("%d\n",ans);
81     }
82   }
83   return 0;
84 }
View Code

 

 

The article write by nealgavin