刷题总结——game（hdu4616）

题目：

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

Sample Output

146
158

题解

大意：给定一颗树··树上节点有不同价值的礼物···有些节点有陷阱··现可以选择从任意一个节点出发按照任意路线走···但不能返回··规定碰到一定数量的陷阱时就会停止走动··问最多可以拿到多少价值总和的礼物？

写了2个多小时发现有处细节错了··就是一旦碰到规定的陷阱数C就必须停止··然而我的dp并没有注意到这一点··导致整个代码方程直接错了···

参照了网上的方法：http://blog.csdn.net/martinue/article/details/51025232

另外这道题还发现了我的一个误区：求一颗树中的最长链的相关问题··如果我们要确定每个节点上的最长链··也就是说如果问到一个节点我们就能答出它的最长链的话,我们需要按照上一篇文章Bob‘s race那样dp··但遇到类似于这道题··如果只是询问最长链而并不需要知道节点与最长链的对应关系时··我们可以按照这道题的方式进行Dp,详见代码·差别很明显

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<string>
#include<cstring>
using namespace std;
const int N=50005;
const int inf=0x3f3f3f3f;
int first[N],nxt[N*2],go[N*2],tot,f[N][5][2],val[N],n,T,C,ans;
bool jud[N];
inline int R()
{
  char c;int f=0;
  for(c=getchar();c<'0'||c>'9';c=getchar());
  for(;c<='9'&&c>='0';c=getchar())  f=(f<<3)+(f<<1)+c-'0';
  return f;
}
inline void comb(int a,int b)
{
  nxt[++tot]=first[a],first[a]=tot,go[tot]=b;
  nxt[++tot]=first[b],first[b]=tot,go[tot]=a;
}
inline void pre()
{
  tot=ans=0;
  memset(first,0,sizeof(first));memset(f,-inf,sizeof(f));
}
inline void dfs(int u,int fa)
{
  f[u][jud[u]][jud[u]]=val[u];ans=max(ans,val[u]);
  for(int e=first[u];e;e=nxt[e]) 
  {
    int v=go[e];if(v==fa)  continue;
    dfs(v,u);
    for(int i=0;i<=C;i++)
      for(int j=0;j+i<=C;j++)
      {
        if(j+i<=C)  ans=max(ans,f[u][i][1]+f[v][j][1]);
        if(j+i<C)   ans=max(ans,f[u][i][0]+f[v][j][0]);
        if(j!=C)    ans=max(ans,f[u][i][1]+f[v][j][0]);
        if(i!=C)    ans=max(ans,f[u][i][0]+f[v][j][1]);
      }
    for(int i=0;i<C;i++)  
    {  
      f[u][i+jud[u]][0]=max(f[u][i+jud[u]][0],f[v][i][0]+val[u]);
      f[u][i+jud[u]][1]=max(f[u][i+jud[u]][1],f[v][i][1]+val[u]);
    }
    if(!jud[u])  f[u][C][1]=max(f[u][C][1],f[v][C][1]+val[u]);
  }
}
int main()
{
 // freopen("a.in","r",stdin);
  T=R();int a,b;
  while(T--)
  {
    n=R(),C=R();pre();
    for(int i=1;i<=n;i++)  val[i]=R(),jud[i]=R();
    for(int i=1;i<n;i++)  a=R(),b=R(),comb(a+1,b+1);
    dfs(1,0);cout<<ans<<endl;
  }
  return 0;
}