鏈接：https://ac.nowcoder.com/acm/contest/888/E
來源：牛客網
Explorer
時間限制：C/C++ 2秒，其他語言4秒
空間限制：C/C++ 524288K，其他語言1048576K
64bit IO Format: %lld
題目描述
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.
輸入描述:
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
輸出描述:
Print a non-negative integer in a single line, denoting the number of valid sizes.
示例1
輸入
5 5 1 2 1 4 2 3 1 2 3 5 2 4 2 4 1 3 4 5 3 4
輸出
2
說明
There are 2 valid sizes : 2 and 3.
For size 2, there exists a path 1→2→3→51 \rightarrow 2 \rightarrow 3 \rightarrow 51→2→3→5.
For size 3, there exists a path 1→2→4→51 \rightarrow 2 \rightarrow 4 \rightarrow 51→2→4→5.
題目大意：給了一個無向圖，n個點，m條邊，每一條邊有一個上下限設位高度，有多少個高度和以從
1到n，可以通過一條邊的條件爲這個高度在這條邊的範圍內。
解題思路：由於區間範圍較大，所以需要對區間離散化，對於一些邊，他能影響到的區間我們可以分爲logn個區間。
我們將這些邊加入到這些區間。然後我們可以遍歷一遍我們的線段樹，對於當前的區間，我們用並查集維護一下連通性。
如果當前的1-n是聯通的話，當前的區間就可以作爲答案。但是這裏的並查集要撤銷，所以不能路徑壓縮。
同時合併的時候要按秩合併。
#include<bits/stdc++.h>
using namespace std;
#define pb(x) push_back(x)
typedef long long ll;
const int mmax = 1e5+5;
const int N = 2e5+5;
int fa[N],hi[N];
struct node
{
int x,y,l,r;
}a[N];
int n;
typedef pair<int,int> pii;
#define mp(x,y) make_pair(x,y)
vector<int>t[N*4];
int Find(int x)
{
while(x!=fa[x]) x = fa[x];
return x;
}
void upd(int rt,int l,int r,int ql,int qr,int k)
{
if(l>=ql && r<=qr)
{
t[rt].pb(k);
return ;
}
int m = (l+r)>>1;
if(ql <= m) upd(rt<<1,l,m,ql,qr,k);
if(qr > m) upd(rt<<1|1,m+1,r,ql,qr,k);
}
struct rec
{
int l,r,del;
rec(int x,int y,int z)
{
l = x, r = y,del = z;
}
};
int ans;
int b[N];
void ask(int rt,int l,int r)
{
vector<rec>tmp;
for(int i=0; i<t[rt].size(); i++)
{
int id = t[rt][i];
int fx = Find(a[id].x);
int fy = Find(a[id].y);
if(fx==fy)continue;
if(hi[fx] > hi[fy])swap(fx,fy);///hi指的是樹高。
fa[fx] = fy; ///小的合併到大的。
tmp.pb(rec(fx,fy,hi[fy]));
hi[fy] = max(hi[fx]+1,hi[fy]);
}
if(Find(1) == Find(n))
{
ans += b[r+1] - b[l]; ///之前這裏寫的return。但是忘了還原並查集了。。。
}
else if(l<r){
int m = (l+r)>>1;
ask(rt<<1,l,m);
ask(rt<<1|1,m+1,r);
}
for(int i=tmp.size()-1; i>=0; i--)
{
rec now = tmp[i];
fa[now.l] = now.l;
hi[now.r] -= now.del;
}
}
int main()
{
int m;
scanf("%d%d",&n,&m);
for(int i=1; i<=n; i++)fa[i] = i ,hi[i] = 0;
int tot=0;
for(int i=1; i<=m; i++)
{
scanf("%d%d%d%d",&a[i].x,&a[i].y,&a[i].l,&a[i].r);
b[++tot] = a[i].l;
b[++tot] = a[i].r+1; ///r+1，是爲了用若干個葉子節點表示區間。
}
sort(b+1,b+1+tot);
tot = unique(b+1, b+1+tot)-b-1;
for(int i=1; i<=m; i++)
{
int l = lower_bound(b+1,b+1+tot,a[i].l)-b;
int r = lower_bound(b+1,b+1+tot,a[i].r+1)-b;
upd(1,1,tot,l,r-1,i); ///這裏是r-1。
}
ans = 0;
ask(1,1,tot);
cout<<ans<<endl;
}
/*
對於上述離散化區間的方法：
對於每一個點r++是爲了處理方便。
例如：[1,4] [3,10] [1,1].
如果讓r++的話
1 3 4 10
這樣的話4個葉子節點無法表示上述區間。。
r++之後
1 2 3 5 11
這樣的話5個葉子節點可以分別表示5個區間
1：[1,1]
2: [2,2]
3: [3,4]
4: [5,10]
5: [11,11]
當然最後一個其實用不到。
這樣的話上述區間就可以剛好用若干個點表示。
*/