2017多校第3场 HDU 6060 RXD and dividing 思维，构造最优解

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6060

题意：给出一颗n个节点的树，要求将2-n号节点分成k部分，然后再将每一部分加上1号节点，定义每一部分的val为其中的点在原图上的最小斯坦纳树，问总的val最大可能是多少。

解法：官方题解：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000010;
typedef long long LL;
struct FastIO
{
static const int S = 1310720;
int wpos;
char wbuf[S];
FastIO() : wpos(0) {}
inline int xchar()
{
static char buf[S];
static int len = 0, pos = 0;
if (pos == len)
pos = 0, len = fread(buf, 1, S, stdin);
if (pos == len) exit(0);
return buf[pos ++];
}
inline int xuint()
{
int c = xchar(), x = 0;
while (c <= 32) c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x;
}
inline int xint()
{
int s = 1, c = xchar(), x = 0;
while (c <= 32) c = xchar();
if (c == '-') s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline void xstring(char *s)
{
int c = xchar();
while (c <= 32) c = xchar();
for (; c > 32; c = xchar()) * s++ = c;
*s = 0;
}
inline void wchar(int x)
{
if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
wbuf[wpos ++] = x;
}
inline void wint(LL x)
{
if (x < 0) wchar('-'), x = -x;
char s[24];
int n = 0;
while (x || !n) s[n ++] = '0' + x % 10, x /= 10;
while (n--) wchar(s
);
}
inline void wstring(const char *s)
{
while (*s) wchar(*s++);
}
~FastIO()
{
if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
}
} io;
int head[maxn], edgecnt, n, k;
struct edge{
int to,len,next;
}E[maxn*2];
void init(){
memset(head,-1,sizeof(head));
edgecnt=0;
}
void add(int u, int v, int w){
E[edgecnt].to = v, E[edgecnt].len = w, E[edgecnt].next = head[u], head[u] = edgecnt++;
}
int sz[maxn];
LL ans;
void dfs(int x, int fa){
sz[x] = 1;
for(int i = head[x]; ~i; i=E[i].next){
int v = E[i].to;
if(v == fa) continue;
dfs(v, x);
ans += (LL)E[i].len * min(sz[v], k);
sz[x] += sz[v];
}
}
int main()
{
while(1)
{
n = io.xint();
k = io.xint();
init();
for(int i=1; i<n; i++){
int u,v,w;
u = io.xint();
v = io.xint();
w = io.xint();
add(u, v, w);
add(v, u, w);
}
ans = 0;
dfs(1, -1);
printf("%lld\n", ans);
}
return 0;
}