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.
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; }