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

5 5
1 2 1 4
2 3 1 2
3 5 2 4
2 4 1 3
4 5 3 4

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 = 486;
const int maxn = 100086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
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<<2];

void update(int l,int r,int rt,int L,int R,int id){
    if(rt==0){ return;}
    if(L<=l&&R>=r){
        eg[rt].push_back(id);
        return;
    }
    int mid = (l+r)>>1;
    if(L<=mid)update(l,mid,rt*2,L,R,id);
    if(R>mid)update(mid+1,r,rt*2+1,L,R,id);
}

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

int ans = 0;
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],1});
            f[t1]=f[t2];
        }else if(rk[t1]>rk[t2]){
            tmp.push(node{f[t2],1});
            f[t2]=f[t1];
        }else{
            tmp.push(node{f[t2],2});
            f[t2]=f[t1];
            rk[t2]++;
        }
    }
    
    if(l==r){
        if(getf(1)==getf(n)&&l!=tot){
            ans+=rem[r+1]-rem[l];
        }
        while (!tmp.empty()){
            node it = tmp.top();
            tmp.pop();
            f[it.num]=it.num;
            if(it.type==2){
                rk[it.num]--;
            }
        }
        return;
    }
    int mid = (l+r)>>1;
    dfs(lson);
    dfs(rson);
    while (!tmp.empty()){
        node it = tmp.top();
        tmp.pop();
        f[it.num]=it.num;
        if(it.type==2){
            rk[it.num]--;
        }
    }
}

int get_id(int x){
    return lower_bound(rem+1,rem+1+tot,x)-rem;
}

int main() {

    scanf("%d%d",&n,&m);
    for(int i=0;i<=n;i++){
        f[i]=i;
        rk[i]=1;
    }

    for(int i=1;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+1;
    }
    sort(rem+1,rem+1+tot);
    tot = unique(rem+1,rem+1+tot)-rem-1;

    for(int i=1;i<=m;i++){
        cout<<get_id(e[i].l)<<" "<<get_id(e[i].r+1)-1<<endl;
        update(1,tot,1,get_id(e[i].l),get_id(e[i].r+1)-1,i);
    }
    dfs(1,tot,1);
    printf("%d
",ans);
    return 0;
}