poj2486

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我是靠谱客的博主强健白昼，最近开发中收集的这篇文章主要介绍poj2486，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6591		Accepted: 2165

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.

Output

For each test case, output the maximal numbers of apples Wshxzt can eat at a line.

Sample Input

Sample Output

11
2

题意：求用最多用k步采摘最多的苹果数,点u->v算一步,点v->u即回来也算一步

分析：对于以u为根的树，走j步,现在有u->v可走(即v为根的子树),可以从u->a>b...->b->a->u->v->....这样走j步,可以是u->v->....->v->u->a->b->.....这样走j步

这样的话就要知道走j步回到u的最值然后再走向v,和需要知道回到v的最值然后走向u的其他子节点

所以本题用到三维数组dp[u][j][0]表示走j步不回到点u采摘的最多苹果数,dp[u][j][1]表示走j步回到u采摘的最多苹果数

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=200+10;
int n,k,size;
//dp[u][j][0]表示以u为根的树经过j步没有回到点u得到的最值
//dp[u][j][1]表示以u为根的树经过j步回到点u得到的最值
int dp[MAX][MAX][2],val[MAX],head[MAX];
struct Edge{
int v,next;
Edge(){}
Edge(int V,int NEXT):v(V),next(NEXT){}
}edge[MAX*2];
void Init(int num){
for(int i=1;i<=num;++i){
head[i]=-1;
for(int j=1;j<=k;++j)dp[i][j][0]=dp[i][j][1]=0;
}
size=0;
}
void InsertEdge(int u,int v){
edge[size]=Edge(v,head[u]);
head[u]=size++;
}
void dfs(int u,int father){
for(int i=head[u];i != -1;i=edge[i].next){
int v=edge[i].v;
if(v == father)continue;
dfs(v,u);
for(int j=k;j>=0;--j){
for(int t=1;j+t<=k;++t){
dp[u][j+t][0]=max(dp[u][j+t][0],dp[u][j][1]+dp[v][t-1][0]+val[v]);
if(t>=2)dp[u][j+t][0]=max(dp[u][j+t][0],dp[u][j][0]+dp[v][t-2][1]+val[v]);
if(t>=2)dp[u][j+t][1]=max(dp[u][j+t][1],dp[u][j][1]+dp[v][t-2][1]+val[v]);
}
}
}
}
int main(){
int u,v;
while(~scanf("%d%d",&n,&k)){
Init(n);
for(int i=1;i<=n;++i)scanf("%d",&val[i]);
for(int i=1;i<n;++i){
scanf("%d%d",&u,&v);
InsertEdge(u,v);
InsertEdge(v,u);
}
dfs(1,-1);
printf("%dn",dp[1][k][0]+val[1]);
}
return 0;
}