题目链接
题解
倍增的思想很棒
题目实际上就是每次让我们合并两个区间对应位置的数，最后的答案(ans = 9 imes 10^{tot - 1})，(tot)是联通块数，因为要去前导(0)，首位不为(0)即可
如何快速合并两个区间？
倍增！
每次合并两个区间，我们就利用倍增分成(logn)个区间，先用并查集维护其联通性
合并完之后，由大区间推向小区间，将每个倍增的大区间分成两半，分别和其联通块的代表区间的两半合并
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 100005,maxm = 100005,INF = 1000000000,P = 1000000007;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int n,m,cnt,pre[maxn * 20],sta[maxn * 20],f[maxn][20],bin[30];
int qpow(int a,int b){
int ans = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) ans = 1ll * ans * a % P;
return ans;
}
inline int find(int u){return u == pre[u] ? u : pre[u] = find(pre[u]);}
inline void merge(int a,int b){
int fa = find(a),fb = find(b);
pre[max(fa,fb)] = min(fa,fb);
}
int main(){
bin[0] = 1; for (int i = 1; i <= 25; i++) bin[i] = bin[i - 1] << 1;
n = read(); m = read();
for (int j = 0; j <= 17; j++)
for (int i = 1; i <= n; i++){
if (i + bin[j] - 1 > n) break;
f[i][j] = ++cnt; sta[cnt] = i; pre[cnt] = cnt;
}
int l,r,ll,rr,len;
while (m--){
l = read(); r = read(); ll = read(); rr = read();
len = r - l + 1;
for (int i = 17; i >= 0; i--)
if (l + bin[i] - 1 <= r){
merge(f[l][i],f[ll][i]);
l += bin[i]; ll += bin[i];
}
}
int u;
for (int j = 17; j; j--)
for (int i = 1; i <= n; i++){
if (i + bin[j] - 1 > n) break;
u = find(f[i][j]);
if (sta[u] != i){
merge(f[i][j - 1],f[sta[u]][j - 1]);
merge(f[i + bin[j - 1]][j - 1],f[sta[u] + bin[j - 1]][j - 1]);
}
}
int tot = 0;
for (int i = 1; i <= n; i++) if (find(f[i][0]) == f[i][0]) tot++;
printf("%lld
",9ll * qpow(10,tot - 1) % P);
return 0;
}