Apple Tree
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7789 | Accepted: 2606 |
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.
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.
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.
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
题目大意：给出一个n个节点的树，每一个节点上有个值，问不超过k步最高能够获得的值。i到j算一步，j到i也算一步
输入： 输入n和k。然后是n个节点的值，然后是n-1个i j代表了i和j节点相邻。
根是1.
非常easy看出来这是一个树状dp。dp[i][j]代表了以i节点为根。用j步能够得到的最大值，可是由于走到子树算是一步，走回到根也是一步，所以就要有两个dp关系，dp1[i][j]代表从i节点走j步又回到j节点的最大值，dp2[i][j]代表从i节点走j步不会到i节点的最大值。
那么状态转移方程为：当前节点为u。子树为v
回到i节点时：在节点u走j步，在子树v中走k步,从u到v和从v到u共走两步。那么在除v之外的其它子树走了j-k-2步。
dp1[u][j] = max(dp1[u][j],dp1[u][j-k-2]+dp1[v][k])
不回到i节点时：从节点u走j步
1.不在v子树中返回u，那么会在其它子树中返回u。在v中走k步。在u到v走一步，在除v外的子树走j-k-1步。
dp2[u][j] = max(dp2[u][j],dp1[u][j-k-1]+dp2[v][k])
2.在v子树中返回u，那么会在其它子树中存在不返回u的，在v中走k步。在u到v和v到u走两步。在除v之外的子树走j-k-2步。
dp2[u][j] = max(dp2[u][j],dp2[u][j-k-2]+dp1[v][k])
注意1：输入的节点i。j是相邻的关系。根是1。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std ;
struct tree{
int v , next ;
}edge[110];
int head[110] , cnt ;
int dp1[110][210] , dp2[110][210] ;//dp1返回。dp2不返回 dp[i][j]:从i节点出发使用j步能够得到的最大值
int c[110] , n , m ;
void add(int u,int v) {
edge[cnt].v = v ;
edge[cnt].next = head[u] ;
head[u] = cnt++ ;
return ;
}
void dfs(int u) {
int i , j , k , v ;
dp1[u][0] = dp2[u][0] = c[u] ;
if( head[u] == -1 )
return ;
for(i = head[u] ; i != -1 ; i = edge[i].next) {
v = edge[i].v ;
dfs(v) ;
}
for(i = head[u] ; i != -1 ; i = edge[i].next) {
v = edge[i].v ;
for(j = m ; j >= 0 ; j--) {
for(k = 0 ; k <= j ; k++) {
if( k+2 <= j ) {
dp1[u][j] = max(dp1[u][j],dp1[u][j-k-2]+dp1[v][k]) ;
dp2[u][j] = max(dp2[u][j],dp2[u][j-k-2]+dp1[v][k]) ;
}
if( k+1 <= j )
dp2[u][j] = max(dp2[u][j],dp1[u][j-k-1]+dp2[v][k]) ;
}
}
}
}
int Map[110][110] ;
queue <int> que ;
void bfs(int n) {
while( !que.empty() ) que.pop() ;
que.push(1) ;
int i , u , v ;
while( !que.empty() ) {
u = que.front() ;
que.pop() ;
for(i = 1 ; i <= n ; i++){
if( Map[u][i] == 1 ) {
Map[u][i] = Map[i][u] = 0 ;
add(u,i) ;
que.push(i) ;
}
}
}
}
int main() {
int i , j , u , v ;
while( scanf("%d %d", &n, &m) != EOF ) {
memset(head,-1,sizeof(head)) ;
memset(dp1,0,sizeof(dp1)) ;
memset(dp2,0,sizeof(dp2)) ;
memset(Map,0,sizeof(Map)) ;
cnt = 0 ;
for(i = 1 ; i <= n ; i++)
scanf("%d", &c[i]) ;
for(i = 1 ; i < n ; i++) {
scanf("%d %d", &u, &v) ;
Map[u][v] = Map[v][u] = 1 ;
}
bfs(n) ;
dfs(1) ;
int max1 = 0 ;
for(i = 0 ; i <= m ; i++) {
max1 = max(max1,dp2[1][i]) ;
}
printf("%d\n", max1) ;
}
return 0 ;
}