http://acm.hdu.edu.cn/showproblem.php?pid=4313
Problem Description
Machines have once again attacked the kingdom of Xions. The kingdom of Xions has N cities and N-1 bidirectional roads. The road network is such that there is a
unique path between any pair of cities.
Morpheus has the news that K Machines are planning to destroy the whole kingdom. These Machines are initially living in K different cities of the kingdom and
anytime from now they can plan and launch an attack. So he has asked Neo to destroy some of the roads to disrupt the connection among Machines. i.e after destroying those roads there should not be any path between any two Machines.
Since the attack can be at any time from now, Neo has to do this task as fast as possible. Each road in the kingdom takes certain time to get destroyed and they
can be destroyed only one at a time.
You need to write a program that tells Neo the minimum amount of time he will require to disrupt the connection among machines.
Input
The first line is an integer T represents there are T test cases. (0<T <=10)
For each test case the first line input contains two, space-separated integers, N and K. Cities are numbered 0 to N-1. Then follow N-1 lines, each containing three, space-separated integers, x y z, which means there is a bidirectional road connecting city x
and city y, and to destroy this road it takes z units of time.Then follow K lines each containing an integer. The ith integer is the id of city in which ith Machine is currently located.
2 <= N <= 100,000
2 <= K <= N
1 <= time to destroy a road <= 1000,000
Output
For each test case print the minimum time required to disrupt the connection among Machines.
Sample Input
1 5 3 2 1 8 1 0 5 2 4 5 1 3 4 2 4 0
Sample Output
10 Hint Neo can destroy the road connecting city 2 and city 4 of weight 5 , and the road connecting city 0 and city 1 of weight 5. As only one road can be destroyed at a time, the total minimum time taken is 10 units of time. After destroying these roads none of the Machines can reach other Machine via any path.
/**
hdu 4313 并查集+贪心
题目大意：
给定n（n<=100000）个节点的树，每条边有一定的权值，有m个点是危险的，现在想将树分成m块使得每块中恰好只有一个危险的点，问最小的花费是多少。
解题思路：
类似Kruskal的贪心过程，将节点按照从小到大排序，以每个危险的节点为并查集的根节点
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=100005;
struct note
{
int u,v;
LL w;
bool operator < (const note &other) const
{
return w>other.w;
}
} edge[maxn];
int father[maxn];
bool flag[maxn];
int n,m;
void init()
{
for(int i=0; i<=n; i++)
{
father[i]=i;
}
}
int find(int u)
{
if(u==father[u])
return u;
return father[u]=find(father[u]);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0; i<n-1; i++)
{
int u,v;
LL w;
scanf("%d%d%I64d",&edge[i].u,&edge[i].v,&edge[i].w);
}
memset(flag,false,sizeof(flag));
for(int i=0; i<m; i++)
{
int u;
scanf("%d",&u);
flag[u]=true;
}
init();
LL sum=0;
sort(edge,edge+n-1);
for(int i=0; i<n-1; i++)
{
int x=find(edge[i].u);
int y=find(edge[i].v);
if(flag[x]&&flag[y])
{
sum+=edge[i].w;
}
else if(flag[x])
{
father[y]=x;
}
else
{
father[x]=y;
}
}
printf("%I64d\n",sum);
}
return 0;
}
/**
hdu 4313 树形dp
题目大意：
给定n（n<=100000）个节点的树，每条边有一定的权值，有m个点是危险的，现在想将树分成m块使得每块中恰好只有一个危险的点，问最小的花费是多少。
解题思路：
dp[i][0]代表的是在当前以i节点为根节点的子树中，i所在的连通块中没有危险节点的最小花费；
dp[i][1]代表的是在当前以i节点为根节点的子树中，i所在的连通块中有危险节点的最小花费；
如果i是叶子节点：如果i为危险点dp[i][0] = inf，dp[i][1]= 0；否则dp[i][0] = 0,dp[i][1] = inf；
如果i不是叶子节点：如果i是危险点dp[i][0] = inf , dp[i][1] = sigma min（dp[son][0]，dp[son][1]+w）；
否则dp[i][0] = sigma min（dp[son][0]，dp[son][1]+w），dp[i][1] = min（dp[i][0] – min（dp[son][0]，dp[son][1]+w）+dp[son][1]）。
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn=100005;
const LL inf=0x3fffffffffffffffLL;
int head[maxn],ip;
LL dp[maxn][2];
int n,m;
bool flag[maxn];
struct note
{
int v,next;
LL w;
} edge[maxn*2];
void init()
{
memset(head,-1,sizeof(head));
ip=0;
}
void addedge(int u,int v,LL w)
{
edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++;
}
void dfs(int u,int pre)
{
/// printf("%d->%d\n",pre,u);
if(flag[u])
{
dp[u][0]=inf;
dp[u][1]=0;
for(int i=head[u]; i!=-1; i=edge[i].next)
{
int v=edge[i].v;
LL w=edge[i].w;
if(v==pre)continue;
dfs(v,u);
dp[u][1]+=min(dp[v][0],dp[v][1]+w);
}
return;
}
else
{
dp[u][0]=0;
dp[u][1]=inf;
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
LL w=edge[i].w;
if(v==pre) continue;
dfs(v,u);
dp[u][0]+=min(dp[v][0],dp[v][1]+w);
}
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
LL w=edge[i].w;
if(v==pre)continue;
dp[u][1]=min(dp[u][1],dp[u][0]-min(dp[v][0],dp[v][1]+w)+dp[v][1]);
}
return;
}
}
/**void dfs1(int u,int pre)
{
printf("%d->%d\n",pre,u);
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v==pre)continue;
dfs1(v,u);
}
}*/
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
init();
for(int i=1; i<n; i++)
{
int u,v;
LL w;
scanf("%d%d%I64d",&u,&v,&w);
u++;
v++;
addedge(u,v,w);
addedge(v,u,w);
// printf("%d-->%d\n",u,v);
// printf("%d-->%d\n",v,u);
}
memset(flag,false,sizeof(flag));
for(int i=0; i<m; i++)
{
int x;
scanf("%d",&x);
x++;
flag[x]=true;
}
dfs(1,-1);
//dfs1(1,-1);
/**for(int i=1;i<=n;i++)
{
printf("%d %I64d %I64d\n",i,dp[i][0],dp[i][1]);
}*/
printf("%I64d\n",min(dp[1][0],dp[1][1]));
}
return 0;
}