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 = 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\n",ans);return 0;
}

View Code

转载于:https://www.cnblogs.com/ZGQblogs/p/11344044.html