codeforces 592D 树链剖分

题意：给一棵树，边上有权值，有两种操作

1 a b f 输出f div（a、b两点间的路径权值和）  向下取整

2 a f  把第a条边的权值修改为f~~~

一开始就想到是树链剖分，但是被数据范围给吓到了，开了一个long double，，但是long double卡效率啊~~结果TLE

后来改为long long,特判一下如果大于1e18就直接赋值为1e18~~~避免溢出  AC了~~   不能用long double声明变量啊~~~~

结果我去看别人的代码，有更加简单的方法~~

由于边的大小只会越来越小，一直小到1，可以利用这一点做文章~~

我们可以暴力往上搜索，且一路合并路径为1的边，（即更改fa[u]的值）

如son->father->grandfather这一条路径~~

当处理到son这个节点时，如果father->grandfather这条边的大小为1的话，那么可以直接把fa[son]=grandfather,跳过father~~~直接处理grandfather这个点~~~

#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 ll long long#define INF 0x3f3f3f3f3f3f3f3f#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=400010;  //修改啊const int cnt_v=200010;int head[cnt_v],cnt_e;struct EDGE{int u,v,next;int cost;}edge[cnt_edge];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++;}void init(){cnt_e=0;memset(head,-1,sizeof(head));}#define erg(i,u) for(int i=head[u];i!=-1;i=edge[i].next)int tot;int lr[cnt_v][2];ll cost[cnt_v];int idx[cnt_v];int fa[cnt_v];void dfs(int u,int pre){    fa[u]=pre;    lr[u][0]=tot;    erg(i,u)    {        int v=edge[i].v;        if(v==pre)continue;        idx[v]=edge[i].cost;//v这个点  对应的是第cost条边~~~~        dfs(v,u);    }    lr[u][1]=tot++;}int bingo(int k)//进行路径压缩，即修改fa的值~~{    return cost[idx[k]]==1?fa[k]=bingo(fa[k]):k;}void solve(int a,int b,ll &c){    if(!a||!c)return;    if(lr[a][0]<=lr[b][0]&&lr[b][1]<=lr[a][1])return;    c/=cost[idx[a]];    solve(fa[a]=bingo(fa[a]),b,c);}int main(){    init();    int n,m;    read(n,m);    for1(i,n-1)    {        int a,b;        read(a,b);read(cost[i]);        addedge(a,b,i);        addedge(b,a,i);    }    tot=1;    dfs(1,0);    int opt,a,b;ll ff;    while(m--)    {        read(opt);        if(opt==1)        {            read(a,b);            read(ff);            solve(a,b,ff);            solve(b,a,ff);            outln(ff);        }        else        {            read(a);read(ff);            cost[a]=ff;        }    }}

树链剖分的代码：

#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 ll long long#define INF 1e18#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=400010;  //Ð޸İ¡const int cnt_v=200010;int head[cnt_v],cnt_e;struct EDGE{int u,v,next,cost;}edge[cnt_edge];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++;}void init(){cnt_e=0;memset(head,-1,sizeof(head));}#define erg(i,u) for(int i=head[u];i!=-1;i=edge[i].next)int top[cnt_v];int fa[cnt_v];int deep[cnt_v];int num[cnt_v];int p[cnt_v];int son[cnt_v];int pos;int eee[cnt_v][2];ll cost[cnt_v];void INIT(){    init();    pos = 0;    memset(son,-1,sizeof(son));}void dfs1(int u,int pre,int d){    deep[u]=d;    fa[u]=pre;    erg(i,u)    {        int v=edge[i].v;        if(v==pre)continue;        dfs1(v,u,d+1);        num[u]+=num[v];        if(son[u]==-1||num[v]>num[son[u]])            son[u]=v;    }}void getpos(int u,int sp){    top[u]=sp;    p[u]=++pos;    if(son[u]==-1)return;    getpos(son[u],sp);    erg(i,u)    {        int v=edge[i].v;        if(v!=son[u]&&v!=fa[u])            getpos(v,v);    }}long double sum[cnt_v*3];#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1void build(int l,int r,int rt){    sum[rt]=0;    if(l==r)return;    int m=l+r>>1;    build(lson);    build(rson);}void modify(int l,int r,int rt,int pos,long double val){    if(l==r)    {        sum[rt]=val;        return;    }    int m=l+r>>1;    if(pos<=m)modify(lson,pos,val);    else modify(rson,pos,val);    if((long double)sum[rt<<1]*sum[rt<<1|1]>=INF)        sum[rt]=INF;    else        sum[rt]=sum[rt<<1]*sum[rt<<1|1];}ll query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)return sum[rt];    int m=l+r>>1;    long double ret=1.0;    if(L<=m)ret*=(long double)query(L,R,lson);    if(R>m)ret*=(long double)query(L,R,rson);    return ret>INF?INF:(ll)ret;}ll find(int u,int v){    int f1=top[u],f2=top[v];    long double ret=1.0;    while(f1!=f2)    {        if(deep[f1]<deep[f2])        {            swap(f1,f2);            swap(u,v);        }        ret*=(long double)query(p[f1],p[u],1,pos,1);        if(ret>=INF)return INF;        u=fa[f1];f1=top[u];    }    if(u==v)return ret;    if(deep[u]>deep[v])swap(u,v);    ret*=(long double)query(p[son[u]],p[v],1,pos,1);    return ret>=INF?INF:ret;}int main(){    int n,m;    INIT();    read(n,m);    for1(i,n-1)    {        read(eee[i][0],eee[i][1]);        read(cost[i]);        addedge(eee[i][0],eee[i][1]);        addedge(eee[i][1],eee[i][0]);    }    dfs1(1,0,0);    getpos(1,1);    build(1,pos,1);    for(int i=1;i<n;i++)    {        if(deep[eee[i][0]]>deep[eee[i][1]])            swap(eee[i][0],eee[i][1]);        modify(1,pos,1,p[eee[i][1]],cost[i]);    }    int opt,a,b;    ll  ff,gg;    while(m--)    {        read(opt);        if(opt==1)        {            read(a,b);            read(ff);            if(a==b)            {                outln(ff);                continue;            }            gg=find(a,b);            ff/=gg;            outln(ff);        }        else        {            read(a);            read(ff);            modify(1,pos,1,p[eee[a][1]],ff);        }    }}