BZOJ3784 树分治、RMQ~~

Description

给定一个N个结点的树，结点用正整数1..N编号。每条边有一个正整数权值。用d(a,b）表示从结点a到结点b路边上经过边的权值。其中要求a<b.将这n*(n-1)/2个距离从大到小排序，输出前M个距离值。

http://blog.csdn.net/u012915516/article/details/48914879  和超级钢琴家一样的思路。可以先参见这道题~~

跑一次树分治，可以得到不同情况下某点的dist值，以及该情况下该点可以到达的子树区间~~

开一个优先队列，存放四元数据(i,l,r,t):=某种状态下某点对应的下标i,该情况下该点可以到达的区间,以及这个区间内最大的下标值~~~RMQ~~

每次取出队列顶点之后,如果[l,t-1]、[t+1,r]可以形成正确的数据，还要添加到队列里~~

#include <algorithm>#include <iostream>#include<string.h>#include <fstream>#include <math.h>#include <vector>#include <cstdio>#include <string>#include <queue>#include <stack>#include <map>#include <set>#define exp 1e-8#define fi first#define se second#define ll long long#define INF 0x3f3f3f3f#define pb(a) push_back(a)#define mp(a,b) make_pair(a,b)#define all(a) a.begin(),a.end()#define mm(a,b) memset(a,b,sizeof(a));#define for0(a,b) for(int a=0;a<=b;a++)//0---(b-1)#define for1(a,b) for(int a=1;a<=b;a++)//1---(b)#define rep(a,b,c) for(int a=b;a<=c;a++)//b---c#define repp(a,b,c)for(int a=b;a>=c;a--)///#define cnt_one(i) __builtin_popcount(i)#define stl(c,itr) for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)using namespace std;void bug(string m="here"){cout<<m<<endl;}template<typename __ll> inline void READ(__ll &m){__ll x=0,f=1;char ch=getchar();while(!(ch>='0'&&ch<='9')){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}m=x*f;}template<typename __ll>inline void read(__ll &m){READ(m);}template<typename __ll>inline void read(__ll &m,__ll &a){READ(m);READ(a);}template<typename __ll>inline void read(__ll &m,__ll &a,__ll &b){READ(m);READ(a);READ(b);}template<typename __ll>inline void read(__ll &m,__ll &a,__ll &b,__ll &c){READ(m);READ(a);READ(b);READ(c);}template<typename __ll>inline void read(__ll &m,__ll &a,__ll &b,__ll &c,__ll &d){READ(m);READ(a);READ(b);READ(c);read(d);}template < class T > inline  void out(T a){if(a<0){putchar('-');a=-a;}if(a>9)out(a/10);putchar(a%10+'0');}template < class T > inline  void outln(T a){out(a);puts("");}template < class T > inline  void out(T a,T b){out(a);putchar(' ');out(b);}template < class T > inline  void outln(T a,T b){out(a);putchar(' ');outln(b);}template < class T > inline  void out(T a,T b,T c){out(a);putchar(' ');out(b);putchar(' ');out(c);}template < class T > inline  void outln(T a,T b,T c){out(a);putchar(' ');outln(b);putchar(' ');outln(b);}template < class T > T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }template < class T > T lcm(T a, T b) { return a / gcd(a, b) * b; }template < class T > inline void rmin(T &a, const T &b) { if(a > b) a = b; }template < class T > inline void rmax(T &a, const T &b) { if(a < b) a = b; }template < class T > T pow(T a, T b) { T r = 1; while(b > 0) { if(b & 1) r = r * a; a = a * a; b /= 2; } return r; }template < class T > T pow(T a, T b, T mod) { T r = 1; while(b > 0) { if(b & 1) r = r * a % mod; a = a * a % mod; b /= 2; } return r; }const int cnt_edge=100100;  //修改啊const int cnt_v=50100;int head[cnt_v],cnt_e;struct EDGE{int u,v,next,cost;}edge[cnt_edge];void init(){cnt_e=0;memset(head,-1,sizeof(head));}void addedge(int u,int v,int cost=0){edge[cnt_e].u=u;edge[cnt_e].v=v;edge[cnt_e].cost=cost;edge[cnt_e].next=head[u];head[u]=cnt_e++;}int n,k;int root,tot;int minn,size[50100];///minn：衡量某个节点是否能当重心，minn越小 就越可以当重心、size[i] i子树的大小bool del[50100];///记录是否已经删除这个点...int l,r,cnt;int dist[2000005];pair<int,int>p[2000005];#define dat(i,l,r,t) DAT{i,l,r,t}struct DAT{    int i,l,r,t;    bool operator < (const DAT &rhs)const    {        return dist[i]+dist[t]<dist[rhs.i]+dist[rhs.t];    }};priority_queue<DAT>que;void getroot(int u,int fa){    size[u]=1;    int maxn=0;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==fa||del[v])continue;///这里必须要加上del[v]        getroot(v,u);        size[u]+=size[v];        maxn=max(maxn,size[v]);    }    maxn=max(maxn,tot-size[u]);    if(maxn<minn)root=u,minn=maxn;}int dfs(int u,int fa,int DIST){    dist[++cnt]=DIST,p[cnt]=mp(l,r);    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(v==fa||del[v])continue;        dfs(v,u,DIST+edge[i].cost);    }}void work(int u){    del[u]=1;    dist[++cnt]=0;p[cnt]=mp(0,0);    l=r=cnt;    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(del[v])continue;        dfs(v,u,edge[i].cost);        r=cnt;    }    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(del[v])continue;        minn=INF,tot=size[v];        getroot(v,u);        work(root);    }}int f[2000005][20];void rmq_init(){    int n=cnt;    for(int i=1;i<=n;i++)f[i][0]=i;    for(int j=1;(1<<j)<=n;j++)        for(int i=1;i+(1<<j)-1<=n;i++)        {            int x=f[i][j-1],y=f[i+(1<<(j-1))][j-1];            f[i][j]=dist[x]>dist[y]?x:y;        }}int rmq(int l,int r){    int k=0;    while((1<<(k+1))<=r-l+1)k++;    int x=f[l][k],y=f[r-(1<<k)+1][k];    return dist[x]>dist[y]?x:y;}int main(){    read(n,k);    init();    for1(i,n-1)    {        int a,b,c;        read(a,b,c);        addedge(a,b,c);        addedge(b,a,c);    }    memset(del,0,sizeof del); ///做题初始化    minn=INF,tot=n;    getroot(1,0);      ///找到一棵树的重，tot为该树的大小....    work(root);    rmq_init();    for1(i,cnt)    {        if(p[i].fi==0)continue;///并不能到达任何点~~        que.push(dat(i,p[i].fi,p[i].se,rmq(p[i].fi,p[i].se)));    }    while(k--)    {        DAT cur=que.top();que.pop();        printf("%d\n",dist[cur.i]+dist[cur.t]);        if(cur.t-1>=cur.l)que.push(dat(cur.i,cur.l,cur.t-1,rmq(cur.l,cur.t-1)));        if(cur.t+1<=cur.r)que.push(dat(cur.i,cur.t+1,cur.r,rmq(cur.t+1,cur.r)));    }}