CF833D Red-Black Cobweb 点分治、树状数组

标签：ack   ==   stdout   之间   ++   ios   不难   需要   com   

传送门

统计所有路径的边权乘积的乘积，不难想到点分治求解。

边权颜色比例在\([\frac{1}{2},2]\)之间，等价于\(2B \geq R , 2R \geq B\)（\(R,B\)表示红色和黑色的边的条数）

所以我们可以在统计的时候，先把所有可能的路径全部乘进答案，然后除掉满足\(2B < R\)或者\(2R < B\)的路径的乘积。显然对于一条路径，这两个条件至多满足一个。

对于两条路径，它们红色、黑色的边数分别为\(B_1,R_1\)和\(B_2,R_2\)，那么需要统计的就是\(R_1 - 2B_1 > 2B_2 - R_2\)或者\(B_1 - 2R_1 > 2R_2 - B_2\)的路径的信息。可以使用树状数组维护。

#include<iostream>
#include<cstdio>
//This code is written by Itst
using namespace std;

inline int read(){
    int a = 0;
    char c = getchar();
    while(!isdigit(c))
        c = getchar();
    while(isdigit(c)){
        a = a * 10 + c - 48;
        c = getchar();
    }
    return a;
}

#define PII pair < int , int >
#define st first
#define nd second
const int MAXN = 1e5 + 7 , MOD = 1e9 + 7;

inline int poww(long long a , int b){
    int times = 1;
    while(b){
        if(b & 1) times = times * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return times;
}

struct Edge{int end , upEd , w , col;}Ed[MAXN << 1];
int head[MAXN] , N , nowSz , minSz , minInd , sum , ans , cnt , cntEd;
bool vis[MAXN];

struct BIT{
#define lowbit(x) ((x) & -(x))
    int BIT0[MAXN << 2] , BIT1[MAXN << 2];
    BIT(){fill(BIT0 , BIT0 + (MAXN << 2) , 1);}
    void add(int pos , int w , int tp){
        pos += 2 * N + 1;
        if(tp == -1) w = poww(w , MOD - 2);
        while(pos <= (N + 1) << 2){
            BIT0[pos] = 1ll * BIT0[pos] * w % MOD;
            BIT1[pos] += tp;
            pos += lowbit(pos);
        }
    }
    PII get(int pos){
        pos += 2 * N + 1;
        int tms = 1 , sum = 0;
        while(pos){
            tms = 1ll * tms * BIT0[pos] % MOD;
            sum += BIT1[pos];
            pos -= lowbit(pos);
        }
        return PII(tms , sum);
    }
}BIT1 , BIT2;

inline void addEd(int a , int b , int c , int d){
    Ed[++cntEd] = (Edge){b , head[a] , c , d};
    head[a] = cntEd;
}

void getSz(int x){
    vis[x] = 1; ++nowSz;
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end]) getSz(Ed[i].end);
    vis[x] = 0;
}

int getRt(int x){
    vis[x] = 1;
    int sz = 1 , maxSz = 0;
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end]){
            int t = getRt(Ed[i].end);
            sz += t; maxSz = max(maxSz , t);
        }
    maxSz = max(maxSz , nowSz - sz);
    if(minSz > maxSz){
        minSz = maxSz;
        minInd = x;
    }
    vis[x] = 0;
    return sz;
}

void addNd(int x , int w , int colR , int colB , int tp){
    sum = 1ll * sum * w % MOD; ++cnt;
    BIT1.add(2 * colR - colB , w , tp);
    BIT2.add(2 * colB - colR , w , tp);
    vis[x] = 1;
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end])
            addNd(Ed[i].end , 1ll * w * Ed[i].w % MOD , colR + Ed[i].col , colB + !Ed[i].col , tp);
    vis[x] = 0;
}

void qryNd(int x , int w , int colR , int colB){
    if(2 * colB >= colR && 2 * colR >= colB)
        ans = 1ll * ans * w % MOD;
    ans = 1ll * ans * sum % MOD * poww(w , cnt) % MOD;
    PII p = BIT1.get(colB - 2 * colR - 1) , q = BIT2.get(colR - 2 * colB - 1);
    ans = 1ll * ans * poww(1ll * p.st * q.st % MOD * poww(w , p.nd + q.nd) % MOD , MOD - 2) % MOD;
    vis[x] = 1;
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end])
            qryNd(Ed[i].end , 1ll * w * Ed[i].w % MOD , colR + Ed[i].col , colB + !Ed[i].col);
    vis[x] = 0;
}

void solve(int x){
    nowSz = cnt = 0; sum = 1; minSz = 1e9;
    getSz(x); getRt(x); x = minInd;
    vis[x] = 1;
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end]){
            qryNd(Ed[i].end , Ed[i].w , Ed[i].col , !Ed[i].col);
            addNd(Ed[i].end , Ed[i].w , Ed[i].col , !Ed[i].col , 1);
        }
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end])
            addNd(Ed[i].end , Ed[i].w , Ed[i].col , !Ed[i].col , -1);
    for(int i = head[x] ; i ; i = Ed[i].upEd)
        if(!vis[Ed[i].end])
            solve(Ed[i].end);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    //freopen("out","w",stdout);
#endif
    N = read();
    for(int i = 1 ; i < N ; ++i){
        int a = read() , b = read() , w = read() , col = read();
        addEd(a , b , w , col); addEd(b , a , w , col);
    }
    ans = 1;
    solve(1);
    cout << ans;
    return 0;
}

CF833D Red-Black Cobweb 点分治、树状数组

标签：ack   ==   stdout   之间   ++   ios   不难   需要   com   

原文地址：https://www.cnblogs.com/Itst/p/10549404.html