HDU 4003 Find Metal Mineral (树形dp）

Find Metal Mineral

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 3463    Accepted Submission(s): 1610

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 11 2 11 3 13 1 21 2 11 3 1

Sample Output

32
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

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题意：给出结点数n,起点s，机器人数k，然后n-1行给出相互连接的两个点，还有这条路线的价值，问你最小花费。

题解：

dp[i][j]表示对于以 i 结点为根结点的子树，放 j 个机器人所需要的权值和。
当 j=0 时表示放了一个机器人下去，遍历完结点后又回到 i 结点了。状态转移方程类似背包。

AC代码：

#include<bits/stdc++.h>using namespace std;struct Node{    int now,next,val;}tree[20005];int dp[10005][15];//dp[i][j]表示对于以i结点为根结点的子树，放j个机器人所需要的权值和。int head[10005];int n,s,k,len;void addedge(int x,int y,int w){    tree[len].now = y;    tree[len].val = w;    tree[len].next = head[x];    head[x] = len++;}void dfs(int root,int fa){        for(int i = head[root];i!=-1;i = tree[i].next)    {        int son = tree[i].now;        if(son == fa)        continue;        dfs(son,root);        for(int j = k;j>=0;j--)        {        //先将dp[son][0]放进树中，因为 dp[son][0]是表示用一个机器人去走完所有子树，最后又回到 pos这个节点，所以花费要乘以2            dp[root][j]+=dp[son][0]+2*tree[i].val;            for(int l = 1;l<=j;l++)//再找到更优的,就是分组背包             dp[root][j] = min(dp[root][j],dp[root][j-l]+dp[son][l]+l*tree[i].val);        }    }}int main(){    int i,x,y,w;    while(~scanf("%d%d%d",&n,&s,&k))    {        len = 0;        memset(head,-1,sizeof(head));        memset(dp,0,sizeof(dp));        for(i = 1;i<n;i++)        {            scanf("%d%d%d",&x,&y,&w);            addedge(x,y,w);            addedge(y,x,w);        }        dfs(s,0);        printf("%d\n",dp[s][k]);    }    return 0;} 

TLE代码....用vector去存...

#include<bits/stdc++.h>#include<vector>using namespace std;vector<int> V[20010];//dp[i][j]表示对于以i结点为根结点的子树，放j个机器人所需要的权值和。 int dp[20010][30];int val[20010];int n,s,k;void dfs(int root,int fa){for(int i=0; i< V[root].size();i++){int son=V[root][i];if(son==fa)continue;dfs(son,root);for(int j=k;j>=0;--j){//先将dp[son][0]放进树中，因为dp[son][0]是表示用一个机器人去走完所有子树，最后又回到 pos这个节点，所以花费要乘以2 dp[root][j]+=dp[son][0]+ 2*val[i];for(int l=1;l<=j;l++)//再找到更优的 {dp[root][j] = min(dp[root][j],dp[root][j-1]+dp[son][l]+l*val[i]);}}}}int main(){int x,y;while(~scanf("%d%d%d",&n,&s,&k)){for(int i=0;i<n;i++) V[i].clear();memset(dp,0,sizeof(dp));for(int i=0;i<n-1;i++){scanf("%d%d%d",&x,&y,&val[i]);V[x].push_back(y);V[y].push_back(x);}dfs(s,0);printf("%d\n",dp[s][k]);}return 0;}