【P4149 [IOI2011]Race】点分治

P4149
给一棵树，每条边有权。求一条简单路径，权值和等于 K，且边的数量最小。

输入格式
第一行包含两个整数 n, K。

接下来 n - 1 行，每行包含三个整数，表示一条无向边的两端和权值。

注意点的编号从 0 开始。

输出格式
输出一个整数，表示最小边数量。

如果不存在这样的路径，输出 -1−1。
cnt[i] 代表路径和为 i 需要的最少边数
那么我们update只要加边update就行，
如果Clear就是删一条边的影响
然后去重复影响注意cnt数组要开1000000

/*if you can't see the repayWhy not just work step by steprubbish is relaxedto ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];inline int read()
{int date = 0,m = 1; char ch = 0;while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();if(ch=='-'){m = -1; ch = getchar();}while(ch>='0' && ch<='9'){date = date*10+ch-'0';ch = getchar();}return date*m;
}/*namespace sgt
{#define mid ((l+r)>>1)#undef mid
}*//*int root[MAX_N],cnt,sz;
namespace hjt
{#define mid ((l+r)>>1)struct node{int l,r,maxx;}T[MAX_N*40];#undef mid
}*/
const int MAX_N = 200025;
int p[MAX_N],eid,k,dep[MAX_N],Size[MAX_N],SIZE,ms,root,maxson[MAX_N],ans;
bool vis[MAX_N];
int cnt[1000025];
void init()
{memset(p,-1,sizeof(p));eid = 0;
}
struct edge
{int v,next,dis;
}e[MAX_N<<1];
void add(int u,int v,int dis)
{e[eid].v = v;e[eid].next = p[u];e[eid].dis = dis;p[u] = eid++;
}
void getroot(int x,int fa)
{Size[x] = 1;maxson[x] = 0;for(int i = p[x];i+1;i=e[i].next){int v = e[i].v;if(v==fa||vis[v]) continue;getroot(v,x);Size[x]+=Size[v];if(Size[v]>maxson[x]) maxson[x] = Size[v];}if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE-Size[x];if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getans(int x,int fa,int dis,int now)
{if(dis>k) return;ans = min(ans,cnt[k-dis]+now);for(int i = p[x];i+1;i=e[i].next){int v = e[i].v;if(v==fa||vis[v]) continue;getans(v,x,dis+e[i].dis,now+1);}}
void update(int x,int fa,int dis,int now)
{if(dis>k) return;cnt[dis] = min(cnt[dis],now);for(int i = p[x];i+1;i=e[i].next){int v = e[i].v;if(v==fa||vis[v]) continue;update(v,x,dis+e[i].dis,now+1);}
}
void Clear(int x,int fa,int dis)
{if(dis>k) return;cnt[dis] = 1e9;for(int i = p[x];i+1;i=e[i].next){int v = e[i].v;if(v==fa||vis[v]) continue;Clear(v,x,dis+e[i].dis);}
}
void fenzhi(int x,int ssize)
{cnt[0] = 0;vis[x] = true;for(int i = p[x];i+1;i=e[i].next){int v = e[i].v;if(vis[v]) continue;getans(v,x,e[i].dis,1);update(v,x,e[i].dis,1);}Clear(x,0,0);for(int i = p[x];i+1;i=e[i].next){int v = e[i].v;if(vis[v]) continue;ms = inf;root = 0;SIZE = Size[v]<Size[x]?Size[v]:(ssize-Size[x]);getroot(v,0);fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));}
}
int main()
{//ios::sync_with_stdio(false);//freopen("a.txt","r",stdin);//freopen("b.txt","w",stdout);int n,a,b,c;scanf("%d%d",&n,&k);init();ans = 1e9;for(int i = 1;i<n;++i){scanf("%d%d%d",&a,&b,&c);a++,b++;add(a,b,c);add(b,a,c);}for(int i = 0;i<=k;++i) cnt[i] = 1e9;ms = inf;root = 0;SIZE = n;getroot(1,-1);fenzhi(root,n);if(ans>=1e9) printf("-1\n");else printf("%d\n",ans);//fclose(stdin);//fclose(stdout);//cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;return 0;
}

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