http://poj.org/problem?id=2486
典型的回溯题目：特别是状态方程用三维的来标记是否要走回路。
题意：一颗树，n个点（1-n），n-1条边，每个点上有一个权值，求从1出发，走V步，最多能遍历到的权值
思路：
树形dp，比较经典的一个树形dp。首先很容易就可以想到用dp[root][k]表示以root为根的子树中最多走k时所能获得的最多苹果数，接下去我们很习惯地会想到将k步在root的所有子结点中分配，也就是进行一次背包，就可以得出此时状态的最优解了，但是这里还有一个问题，那就是在进行背包的时候，对于某个孩子son走完之后是否回到根结点会对后面是否还能分配有影响，为了解决这个问题，我们只需要在状态中增加一维就可以了，用dp[root][k][0]表示在子树root中最多走k步，最后还是回到root处的最大值，dp[root][k][1]表示在子树root中最多走k步，最后不回到root处的最大值。由此就可以得出状态转移方程了：
Apple Tree
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6836 | Accepted: 2268 |
Description
Wshxzt is a lovely girl. She likes apple very much. One
day HX takes her to an apple tree. There are N nodes in the tree. Each node has
an amount of apples. Wshxzt starts her happy trip at one node. She can eat up
all the apples in the nodes she reaches. HX is a kind guy. He knows that eating
too many can make the lovely girl become fat. So he doesn’t allow Wshxzt to go
more than K steps in the tree. It costs one step when she goes from one node to
another adjacent node. Wshxzt likes apple very much. So she wants to eat as many
as she can. Can you tell how many apples she can eat in at most K steps.
Input
There are several test cases in the input
Each test case contains three parts. The first part is two numbers N K,
whose meanings we have talked about just now. We denote the nodes by 1 2 ... N.
Since it is a tree, each node can reach any other in only one route.
(1<=N<=100, 0<=K<=200) The second part contains N integers
(All integers are nonnegative and not bigger than 1000). The ith number is the
amount of apples in Node i. The
third part contains N-1 line. There are two numbers A,B in each line, meaning
that Node A and Node B are adjacent. Input will be ended by the end of
file.
Note: Wshxzt starts at Node 1.
Note: Wshxzt starts at Node 1.
Output
For each test case, output the maximal numbers of
apples Wshxzt can eat at a line.
Sample Input
2 1 0 11 1 2 3 2 0 1 2 1 2 1 3
Sample Output
11 2
dp[root][j][0] = MAX (dp[root][j][0] , dp[root][j-k][0] + dp[son][k-2][0]);//从s出发，要回到s，需要多走两步s-t,t-s,分配给t子树k步，其他子树j-k步，都返回
dp[root][j]][1] = MAX( dp[root][j][1] , dp[root][j-k][0] + dp[son][k-1][1]) ;//先遍历s的其他子树，回到s，遍历t子树，在当前子树t不返回，多走一步
dp[root][j][1] = MAX (dp[root][j][1] , dp[root][j-k][1] + dp[son][k-2][0]);//不回到s（去s的其他子树），在t子树返回，同样有多出两步
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//（1）dp[i][j+2][0] = max(dp[i][j+2][0], dp[i][j-k][0]+dp[son][k][0]);//（2）dp[i][j+1][1] = max(dp[i][j+1][1], dp[i][j-k][0]+dp[son][k][1]); 人留在i的子节点son的子树中//（3）dp[i][j+2][1] = max(dp[i][j+2][1], dp[i][j-k][1]+dp[son][k][0]); 人留在不是son的i的子节点的子树中#include<iostream>#include<cstring>#include<cstdio>using
namespace std;
int dp[300][300][3],head[300],vis[300],w[300];
int len,n,k;
struct
node
{ int
now,next;
} tree[505];int
max(int
x,int y)
{ if(x>y)
return
x;
else
return
y;
}void
add(int
x,int y)
{ tree[len].now = y;
tree[len].next = head[x];
head[x] = len++;
}void
dfs(int
root,int
mark)
{ int
j,son,t,i;
for(i=0;i<=k;i++)
dp[root][i][0] = dp[root][i][1] = w[root];
for(i=head[root];i!=-1;i=tree[i].next)
{
printf("i=%d\n",i);
son = tree[i].now;
if(son == mark)//已经加了，就不要加，不然就死循环。
continue;
dfs(son,root);
for(j = k; j>=1; j--)
{
for(t = 1; t<=j; t++)
{
dp[root][j][1]=max(dp[root][j][1],dp[root][j-t][1]+dp[son][t-2][1]);
dp[root][j][0]=max(dp[root][j][0],dp[root][j-t][1]+dp[son][t-1][0]);
dp[root][j][0]=max(dp[root][j][0],dp[root][j-t][0]+dp[son][t-2][1]);
}
}
}
}int
main()
{ int
i,a,b,j;
while(~scanf("%d%d",&n,&k))
{
len=0;
memset(head,-1,sizeof(head));
memset(dp,0,sizeof(dp));
memset(vis,0,sizeof(vis));
memset(w,0,sizeof(w));
for(i = 1; i<=n; i++)
{
scanf("%d",&w[i]);
}
for(i=1;i<n;i++)
{
cin>>a>>b;
add(a,b);
add(b,a);
}
dfs(1,0);
printf("%d\n",max(dp[1][k][0],dp[1][k][1]));
}
return
0;
} |