[bzoj4569] [SCOI2016] Meng Meng da

Main topic

There is an n-digit number with no leading 0, and there are m-constrained shapes such as [L1,R1]=[L2,R2], which ask the number of satisfying conditions, the answer modulo 10^9+7.

Double

We use ST table, f[i,j] to denote [i,i+2^j-1] this paragraph.
Then each paragraph is initially made into a single set.
For a limit, you can split the log copies and then merge the collections.
Then, if any f[s,t] and f[i,j] belong to the same set, then F[s,t-1] and f[i,j-1] and f[s+2^ (t-1) -1,t-1] and f[i+2^ (j-1) -1,j-1] should all belong to the same collection.
In order to meet this limit, as long as the last layer to do, the next layer of the merger can be.
Attention must be put into the number of large numbers in the number of small.
The statistical answer is very simple, the specific look at the code.

#include <cstdio>#include <algorithm>#define FO (i,a,b) for (i=a;i<=b;i++)#define FD (I,A,B) for (i=a;i>=b;i--)using namespace Std;typedef Long LongllConst intmaxn=100000+Ten, mo=1000000007, maxd= -;intfa[maxn* +],ID[MAXN] [ +],di[maxn* +][2],two[ +];intI,j,k,l1,r1,l2,r2,s,t,n,m,ans,tot;intGETFA (intx) {returnFA[X]?FA[X]=GETFA (Fa[x]): x;}intMergeintXintY) {X=GETFA (x); Y=GETFA (y);if(x>y) Swap (x, y);if(x!=y) fa[y]=x;}intMain () {two[0]=1; Fo (i,1, Maxd) two[i]=two[i-1]*2; scanf"%d%d", &n,&m);if(n==1) {printf ("10\n");return 0; } FO (i,1, N) Fo (J,0, Maxd)IDI [j]=++tot,di[tot][0]=i,di[tot][1]=j; Fo (i,1, m) {scanf ("%d%d%d%d", &AMP;L1,&AMP;R1,&AMP;L2,&AMP;R2); FD (J,maxd,0)if(l1+two[j]-1&LT;=R1) {Merge (ID[L1] [j],ID[L2]                [j]);            L1+=TWO[J];L2+=TWO[J]; }} FD (J,maxd,1) Fo (i,1, n) {K=GETFA (IDI            [j]); s=di[k][0];t=di[k][1]; MergeID[S] [t1],IDI [J-1]); MergeID[s+two[t-1]][t-1],ID[i+two[j-1]][j-1]); } ans=1; Fo (i,1, N)if(!fa[IDI [0]]) ans= (LL) ans* (i==1)?9:Ten)%mo; printf"%d\n", ans);}

[bzoj4569] [SCOI2016] Meng Meng da