hdu 4003 树形dp+分组背包

Find Metal Mineral

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)

Total Submission(s): 2106    Accepted Submission(s): 953


[align=left]Problem Description[/align]
Humans have discovered a kind of new metal mineral on Mars which are distributed in point‐like with paths connecting each of them which formed a tree. Now Humans launches k robots on Mars to collect them, and due to the unknown reasons,
the landing site S of all robots is identified in advanced, in other word, all robot should start their job at point S. Each robot can return to Earth anywhere, and of course they cannot go back to Mars. We have research the information of all paths on Mars,
including its two endpoints x, y and energy cost w. To reduce the total energy cost, we should make a optimal plan which cost minimal energy cost.

 

[align=left]Input[/align]
There are multiple cases in the input.

In each case:

The first line specifies three integers N, S, K specifying the numbers of metal mineral, landing site and the number of robots.

The next n‐1 lines will give three integers x, y, w in each line specifying there is a path connected point x and y which should cost w.

1<=N<=10000, 1<=S<=N, 1<=k<=10, 1<=x, y<=N, 1<=w<=10000.
 

[align=left]Output[/align]
For each cases output one line with the minimal energy cost.

 

[align=left]Sample Input[/align]

3 1 1
1 2 1
1 3 1
3 1 2
1 2 1
1 3 1

 

[align=left]Sample Output[/align]

3
2

HintIn the first case: 1->2->1->3 the cost is 3;
In the second case: 1->2; 1->3 the cost is 2;

 

题意：给定一棵树，起点，机器人个数，求遍历所有节点的最小路径和。

ddp[u][i]表示访问结束后以u为根节点的子树上机器人个数，假如有0个机器人，那么肯定是一个机器人去访问完后又返回了，因此需要特殊考虑，

代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-6 16:15:10
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 1000000
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=40009;
int head[maxn],tol;
struct node{
int next,to,val;
node(){};
node(int _next,int _to,int _val):next(_next),to(_to),val(_val){}
}edge[5*maxn];
void add(int u,int v,int val){
edge[tol]=node(head[u],v,val);
head[u]=tol++;
}
int K,dp[maxn][13];
void dfs(int u,int fa){
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(v==fa)continue;
dfs(v,u);
for(int j=K;j>=0;j--){
dp[u][j]+=dp[v][0]+2*edge[i].val;
for(int k=0;k<=j;k++)
dp[u][j]=min(dp[u][j],dp[u][j-k]+dp[v][k]+k*edge[i].val);
}
}
}
int main(){
int i,j,k,m,n,s;
while(~scanf("%d%d%d",&n,&s,&K)){
memset(head,-1,sizeof(head));tol=0;
for(i=1;i<n;i++){
scanf("%d%d%d",&j,&k,&m);
add(j,k,m);
add(k,j,m);
}
memset(dp,0,sizeof(dp));
dfs(s,-1);
printf("%d\n",dp[s][K]);
}
}