hdu 5242——Game——————【树链剖分思想】
Appoint description:
Description
It is well known that Keima Katsuragi is The Capturing God because of his exceptional skills and experience in ''capturing'' virtual girls in gal games. He is able to play $k$ games simultaneously.
One day he gets a new gal game named ''XX island''. There are $n$ scenes in that game, and one scene will be transformed to different scenes by choosing different options while playing the game. All the scenes form a structure like a rooted tree such that the root is exactly the opening scene while leaves are all the ending scenes. Each scene has a value , and we use $w_i$ as the value of the $i$-th scene. Once Katsuragi entering some new scene, he will get the value of that scene. However, even if Katsuragi enters some scenes for more than once, he will get $w_i$ for only once.
For his outstanding ability in playing gal games, Katsuragi is able to play the game $k$ times simultaneously. Now you are asked to calculate the maximum total value he will get by playing that game for $k$ times.
One day he gets a new gal game named ''XX island''. There are $n$ scenes in that game, and one scene will be transformed to different scenes by choosing different options while playing the game. All the scenes form a structure like a rooted tree such that the root is exactly the opening scene while leaves are all the ending scenes. Each scene has a value , and we use $w_i$ as the value of the $i$-th scene. Once Katsuragi entering some new scene, he will get the value of that scene. However, even if Katsuragi enters some scenes for more than once, he will get $w_i$ for only once.
For his outstanding ability in playing gal games, Katsuragi is able to play the game $k$ times simultaneously. Now you are asked to calculate the maximum total value he will get by playing that game for $k$ times.
Input
The first line contains an integer $T$($T \le 20$), denoting the number of test cases.
For each test case, the first line contains two numbers $n, k(1 \le k \le n \le 100000)$, denoting the total number of scenes and the maximum times for Katsuragi to play the game ''XX island''.
The second line contains $n$ non-negative numbers, separated by space. The $i$-th number denotes the value of the $i$-th scene. It is guaranteed that all the values are less than or equal to $2^{31} - 1$.
In the following $n - 1$ lines, each line contains two integers $a, b(1 \le a, b \le n)$, implying we can transform from the $a$-th scene to the $b$-th scene.
We assume the first scene(i.e., the scene with index one) to be the opening scene(i.e., the root of the tree).
For each test case, the first line contains two numbers $n, k(1 \le k \le n \le 100000)$, denoting the total number of scenes and the maximum times for Katsuragi to play the game ''XX island''.
The second line contains $n$ non-negative numbers, separated by space. The $i$-th number denotes the value of the $i$-th scene. It is guaranteed that all the values are less than or equal to $2^{31} - 1$.
In the following $n - 1$ lines, each line contains two integers $a, b(1 \le a, b \le n)$, implying we can transform from the $a$-th scene to the $b$-th scene.
We assume the first scene(i.e., the scene with index one) to be the opening scene(i.e., the root of the tree).
Output
For each test case, output ''Case #t:'' to represent the $t$-th case, and then output the maximum total value Katsuragi will get.
Sample Input
2
5 2
4 3 2 1 1
1 2
1 5
2 3
2 4
5 3
4 3 2 1 1
1 2
1 5
2 3
2 4
Sample Output
Case #1: 10
Case #2: 11
题目大意：给你一棵树，有n个结点，有n-1条边，每个结点都有一个价值，可以从跟到叶子k次，但是只能拿一次价值，问你从根1到达叶子的最大价值。
解题思路：用深搜划分出与根临接的顶点个数数量的子树，回溯的时候从这些子树中可以找出重链，如果在中间有不是重链上的链，则把该链放入优先队列中，否则维护出一条重链。dp[i]表示从叶子到i结点的最大价值和。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
#define LL __int64
const int maxn=1e5+100;
vector<LL>V[maxn];
priority_queue<LL>Q;
LL a[maxn];
LL dp[maxn];
LL dfs(int cur){
int i,v;
LL tmp;
dp[cur]=a[cur];
for(i=0;i<V[cur].size();i++){
v=V[cur][i];
tmp=dfs(v);
if(dp[cur]>tmp+a[cur]){ //如果有其他轻链，把轻链放入队列
Q.push(tmp);
}else{ //如果当前这个链不是最重的链，把从叶子到该结点的这段链放入队列
Q.push(dp[cur]-a[cur]);
dp[cur]=tmp+a[cur];
}
}
return dp[cur];
}
int main(){
int t ,n,m,i,j,k,u,v,cnt=0;
LL ans;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&k);
for(i=1;i<=n;i++)
scanf("%lld",&a[i]);
for(i=1;i<n;i++){
scanf("%d%d",&u,&v);
V[u].push_back(v);
}
dfs(1);
ans=dp[1];
for(i=1;i<k;i++){
ans+=Q.top();
Q.pop();
}
printf("Case #%d: %lld\n",++cnt,ans);
while(!Q.empty())
Q.pop();
for(i=0;i<=n;i++){
V[i].clear();
}
}
return 0;
}
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