标签：des blog http ar io os sp for strong
http://www.spoj.com/problems/FTOUR2/
先前看了一会题解就自己yy出来了。。。对拍过后交tle。。。。。。。。。。。。。。。。。。
自己造了下大数据。。。。。。。。tle。。。。。。。。。。。。。。。。。。。。。。。。。
what？
明早更新吧。。调到吐了。。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }
const int N=200005, oo=~0u>>1;
int ihead[N], cnt, n;
struct dat { int next, to, w; }e[N<<1];
inline void add(int u, int v, int w) {
e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].w=w;
e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].w=w;
}
int root, sz[N], vis[N], ans, h[N], blc[N], g[N], bsz[N], K, mn;
void getroot(int x, int fa, int sum) {
sz[x]=1; int mx=0, y;
rdm(x, i) if(!vis[y=e[i].to] && e[i].to!=fa) {
getroot(y, x, sum);
sz[x]+=sz[y];
mx=max(mx, sz[y]);
}
mx=max(mx, sum-mx);
if(mx<mn) mn=mx, root=x;
}
void cal(int x, int len, int num, int fa) {
h[num]=max(h[num], len); int y;
rdm(x, i) if(!vis[y=e[i].to] && e[i].to!=fa) {
cal(y, len+e[i].w, num+blc[y], x);
}
}
void fix(int len) {
int mx=-oo;
for1(i, 0, len) {
mx=max(mx, g[i]);
g[i]=mx;
}
}
void getbsz(int x, int fa) {
bsz[x]=blc[x];
rdm(x, i) if(!vis[e[i].to] && e[i].to!=fa) getbsz(e[i].to, x), bsz[x]+=bsz[e[i].to];
}
struct QQ {
int dep, id;
bool operator<(const QQ &a) const { return dep<a.dep; }
}q[N];
void dfs(int x, int sum) {
vis[x]=1;
int y, maxdep=-1, kk=K-blc[x];
int tot=0, pos;
rdm(x, i) if(!vis[y=e[i].to]) {
getbsz(y, x);
q[++tot].dep=bsz[y];
q[tot].id=i;
}
sort(q+1, q+1+tot);
for1(i, 1, tot) {
int now=q[i].id, y=e[now].to, s=min(q[i].dep, kk);
for1(j, 0, s) h[j]=-oo;
cal(y, e[now].w, blc[y], x);
if(i==1) for1(j, 0, s) g[j]=h[j];
else {
for1(j, 0, s) {
pos=kk-j; if(pos>maxdep) pos=maxdep;
if(g[pos]!=-oo && h[j]!=-oo) ans=max(g[pos]+h[j], ans);
}
for1(j, 0, s) g[j]=max(g[j], h[j]);
}
maxdep=s;
fix(maxdep);
}
ans=max(ans, g[min(kk, maxdep)]);
rdm(x, i) if(!vis[y=e[i].to]) {
int s=sz[y]>sz[x]?sum-sz[x]:sz[y];
root=0; mn=oo; getroot(y, x, s);
dfs(root, s);
}
}
int main() {
read(n); read(K); int m=getint();
for1(i, 1, m) blc[getint()]=1;
rep(i, n-1) { int u=getint(), v=getint(), w=getint(); add(u, v, w); }
mn=oo;
getroot((n+1)>>1, -1, n);
dfs(root, n);
print(ans);
return 0;
}
After the success of 2nd anniversary (take a look at problem FTOUR for more details), this 3rd year, Travel Agent SPOJ goes on with another discount tour.
The tour will be held on ICPC island, a miraculous one on the Pacific Ocean. We list N places (indexed from 1 to N) where the visitors can have a trip. Each road connecting them has an interest value, and this value can be negative (if there is nothing interesting to view there). Simply, these N places along with the roads connecting them form a tree structure. We will choose two places as the departure and destination of the tour.
Since September is the festival season of local inhabitants, some places are extremely crowded (we call them crowded places). Therefore, the organizer of the excursion hopes the tour will visit at most K crowded places (too tiring to visit many of them) and of course, the total number of interesting value should be maximum.
Briefly, you are given a map of N places, an integer K, and M id numbers of crowded place. Please help us to find the optimal tour. Note that we can visit each place only once (or our customers easily feel bored), also the departure and destination places don‘t need to be different.
There is exactly one case. First one line, containing 3 integers N K M, with 1 <= N <= 200000, 0 <=K <= M, 0 <= M <= N.
Next M lines, each line includes an id number of a crowded place.
The last (N - 1) lines describe (N - 1) two-way roads connected N places, form a b i, with a, b is the id of 2 places, and i is its interest value (-10000 <= i <= 10000).
Only one number, the maximum total interest value we can obtain.
Input: 8 2 3 3 5 7 1 3 1 2 3 10 3 4 -2 4 5 -1 5 7 6 5 6 5 4 8 3 Output: 12
标签：des blog http ar io os sp for strong
原文地址：http://www.cnblogs.com/iwtwiioi/p/4172993.html
