2019牛客暑期多校训练营（第八场）E.Explorer

链接：https://ac.nowcoder.com/acm/contest/888/E
来源：牛客网

Gromah and LZR have entered the fifth level. Unlike the first four levels, they should do some moves in this level.

There are nn_{}n vertices and mm_{}m bidirectional roads in this level, each road is in format (u,v,l,r)(u, v, l, r)_{}(u,v,l,r), which means that vertex uu_{}u and vv_{}v are connected by this road, but the sizes of passers should be in interval [l,r][l, r]_{}[l,r]. Since passers with small size are likely to be attacked by other animals and passers with large size may be blocked by some narrow roads.

Moreover, vertex 11_{}1 is the starting point and vertex nn_{}n is the destination. Gromah and LZR should go from vertex 11_{}1 to vertex nn_{}n to enter the next level.

At the beginning of their exploration, they may drink a magic potion to set their sizes to a fixed positive integer. They want to know the number of positive integer sizes that make it possible for them to go from 11_{}1 to nn_{}n.

Please help them to find the number of valid sizes.

输入描述:

The first line contains two positive integers n,mn,m_{}n,m, denoting the number of vertices and roads.

Following m lines each contains four positive integers u,v,l,ru, v, l, r_{}u,v,l,r, denoting a bidirectional road (u,v,l,r)(u, v, l, r)_{}(u,v,l,r).

1≤n,m≤105,1≤u<v≤n,1≤l≤r≤1091 \le n,m \le 10^5, 1 \le u < v \le n, 1 \le l \le r \le 10^91≤n,m≤105,1≤u<v≤n,1≤l≤r≤109

输出描述:

Print a non-negative integer in a single line, denoting the number of valid sizes.

示例1

输入

输出

2

题意：
给定ｍ条边，每条边有一个通过的阈值，问可以从１到n的值有多少个．
思路：
把这些边放入一个点表示区间的线段树里面．
这真是我从未见过的全新套路，在线段树上dfs，相当于枚举权值，由于每一个点代表区间，所以每次就枚举到了一个区间．

枚举之时用并查集判断．

#include<iostream>

#include<algorithm>

#include<vector>

#include<stack>

#include<queue>

#include<map>

#include<set>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<ctime>

#define fuck(x) cerr<<#x<<" = "<<x<<endl;

#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;

#define lson l,mid,ls

#define rson mid+1,r,rs

#define ls (rt<<1)

#define rs ((rt<<1)|1)

using namespace std;

typedef long long ll;

typedef unsigned long long ull;

const int loveisblue = ;

const int maxn = ;

const int maxm = ;

const int inf = 0x3f3f3f3f;

const ll Inf = ;

const int mod = ;

const double eps = 1e-;

const double pi = acos(-);

int n,m;

struct edge{

    int u,v,l,r;

}e[maxn];

int f[maxn],rk[maxn];

int rem[maxn],tot;

vector<int>eg[maxn<<];

void update(int l,int r,int rt,int L,int R,int id){

    if(rt==){ return;}

    if(L<=l&&R>=r){

        eg[rt].push_back(id);

        return;

    }

    int mid = (l+r)>>;

    if(L<=mid)update(l,mid,rt*,L,R,id);

    if(R>mid)update(mid+,r,rt*+,L,R,id);

}

int getf(int x){

    if(x==f[x]){ return x;}

    return getf(f[x]);

}

int ans = ;

struct node{

    int num,type;

};

void dfs(int l,int r,int rt){

    stack<node>tmp;

    for(auto it:eg[rt]){

        int t1 = getf(e[it].u);

        int t2 = getf(e[it].v);

        if(rk[t1]<rk[t2]){

            tmp.push(node{f[t1],});

            f[t1]=f[t2];

        }else if(rk[t1]>rk[t2]){

            tmp.push(node{f[t2],});

            f[t2]=f[t1];

        }else{

            tmp.push(node{f[t2],});

            f[t2]=f[t1];

            rk[t2]++;

        }

    }

    if(l==r){

        if(getf()==getf(n)&&l!=tot){

            ans+=rem[r+]-rem[l];

        }

        while (!tmp.empty()){

            node it = tmp.top();

            tmp.pop();

            f[it.num]=it.num;

            if(it.type==){

                rk[it.num]--;

            }

        }

        return;

    }

    int mid = (l+r)>>;

    dfs(lson);

    dfs(rson);

    while (!tmp.empty()){

        node it = tmp.top();

        tmp.pop();

        f[it.num]=it.num;

        if(it.type==){

            rk[it.num]--;

        }

    }

}

int get_id(int x){

    return lower_bound(rem+,rem++tot,x)-rem;

}

int main() {

    scanf("%d%d",&n,&m);

    for(int i=;i<=n;i++){

        f[i]=i;

        rk[i]=;

    }

    for(int i=;i<=m;i++){

        scanf("%d%d%d%d",&e[i].u,&e[i].v,&e[i].l,&e[i].r);

        rem[++tot] = e[i].l;

        rem[++tot] = e[i].r+;

    }

    sort(rem+,rem++tot);

    tot = unique(rem+,rem++tot)-rem-;

    for(int i=;i<=m;i++){

        cout<<get_id(e[i].l)<<" "<<get_id(e[i].r+)-<<endl;

        update(,tot,,get_id(e[i].l),get_id(e[i].r+)-,i);

    }

    dfs(,tot,);

    printf("%d\n",ans);

    return ;

}