xtu summer individual 6 E

Find Metal Mineral

Time Limit: 1000ms

Memory Limit: 65768KB

This problem will be judged on HDU. Original ID: 4003
64-bit integer IO format: %I64d      Java class name: Main

 

 

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

3
2

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

 

 

解题：传说中的树形dp。。。嗯 待会再讲吧。。。。

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
struct arc {
    int to,w;
};
vector<arc>g[10010];
int dp[10010][15],n,s,k;
void dfs(int u,int fa){
    for(int i = 0; i < g[u].size(); i++){
        int v = g[u][i].to;
        if(v == fa) continue;
        dfs(v,u);
        for(int t = k; t >= 0; t--){
            dp[u][t] += dp[v][0] + 2*g[u][i].w;
            for(int j = 1; j <= t; j++)
                dp[u][t] = min(dp[u][t],dp[u][t-j]+dp[v][j]+j*g[u][i].w);
        }
    }
}
int main(){
    while(~scanf("%d%d%d",&n,&s,&k)){
        for(int i = 0; i <= n; i++)
            g[i].clear();
        for(int i = 1; i < n; i++){
            int u,v,w;
            scanf("%d %d %d",&u,&v,&w);
            g[u].push_back((arc){v,w});
            g[v].push_back((arc){u,w});
        }
        memset(dp,0,sizeof(dp));
        dfs(s,-1);
        printf("%d
",dp[s][k]);
    }
    return 0;
}