【树形dp】Find Metal Mineral

[HDU4003]Find Metal Mineral

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3686 Accepted Submission(s): 1723

Problem Description

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.

Input

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.

Output

For each cases output one line with the minimal energy cost.

Sample Input

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

Sample Output

Hint

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

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

题目大意：给出一颗带边权树，N个节点，根为S。现在有K个机器人要遍历这N个点，求最小花费？

试题分析：依照以前的设计思路，还是dp[N][K]表示在节点i的子树中用去j个机器人状态。

　　　　　那么这样设计可不可行呢？我们来看一下转移方程：

　　　　　　　那么就需要将j=0单列出来讨论　dp[i][0]=sum(dp[i->son][0]+2*Cost[i->son]) 因为只有一个机器人，所以只能它自己走，结果一定。

　　　　　　　　　　　　　j≠0时：　　　　　dp[i][j]=min(dp[i][j],dp[i->son][t]+t*Cost[i->son]+dp[i][j-t]);有多个机器人时，肯定让它们分开跑最优。

　　　　　dp[i][j]每次应该先加上dp[i->son][0]+2*Cost[i->son]，因为可能用一个机器人遍历这颗子树。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN=100001;
const int INF=999999;
int N,S,K;
int u,w,v;
int Next[MAXN*2],Node[MAXN*2],Root[MAXN*2],Cost[MAXN*2];
int dp[MAXN][12];
int cnt;

void addedge(int u,int v,int w){
	cnt++;
	Node[cnt]=v;
	Cost[cnt]=w;
	Next[cnt]=Root[u];
	Root[u]=cnt;
	return ;
} 

void dfs(int x,int fa){
	int cnt=0;
	for(int x1=Root[x];x1;x1=Next[x1]){
		if(Node[x1]==fa) continue;
		dfs(Node[x1],x);cnt++;
	}
	if(!cnt) {
	    for(int i=0;i<=K;i++) dp[x][i]=0;
		return ;
	}
	for(int x1=Root[x];x1;x1=Next[x1]){
		if(Node[x1]==fa) continue;
		int SON=Node[x1];
		for(int k=K;k>=0;k--){
			if(dp[x][k]==-1){
				if(!k) dp[x][0]=dp[SON][0]+Cost[x1]*2;
				else{
					for(int t=1;t<=k;t++)
					    if(dp[x][k]==-1||dp[x][k]>dp[SON][t]+t*Cost[x1])
					        dp[x][k]=dp[SON][t]+t*Cost[x1];
				}
			} 
			else{
				if(!k) dp[x][k]+=(dp[SON][0]+Cost[x1]*2);
				else{
					dp[x][k]+=dp[SON][0]+2*Cost[x1];
					for(int t=1;t<=k;t++){
					    dp[x][k]=min(dp[x][k-t]+dp[SON][t]+t*Cost[x1],dp[x][k]);
					}
				}
			}
		}
	}
	return ;
}

int main(){
	while(scanf("%d%d%d",&N,&S,&K)!=EOF){
		cnt=0;
		memset(Node,0,sizeof(Node));
		memset(Root,0,sizeof(Root));
		memset(Cost,0,sizeof(Cost));
		memset(Next,0,sizeof(Next));
		for(int i=1;i<N;i++){
			u=read(),v=read(),w=read();
		    addedge(u,v,w);
		    addedge(v,u,w);
		}
		memset(dp,-1,sizeof(dp));
		dfs(S,-1);
		printf("%d
",dp[S][K]);
	}
}