HDU 3461 Code Lock (並查集)

Code Lock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 821 Accepted Submission(s): 292

Problem DescriptionA lock you use has a code system to be opened instead of a key. The lock contains a sequence of wheels. Each wheel has the 26 letters of the English alphabet 'a' through 'z', in order. If you move a wheel up, the letter it shows changes
to the next letter in the English alphabet (if it was showing the last letter 'z', then it changes to 'a').
At each operation, you are only allowed to move some specific subsequence of contiguous wheels up. This has the same effect of moving each of the wheels up within the subsequence.
If a lock can change to another after a sequence of operations, we regard them as same lock. Find out how many different locks exist?

InputThere are several test cases in the input.

Each test case begin with two integers N (1<=N<=10000000) and M (0<=M<=1000) indicating the length of the code system and the number of legal operations.

Then M lines follows. Each line contains two integer L and R (1<=L<=R<=N), means an interval [L, R], each time you can choose one interval, move all of the wheels in this interval up.

The input terminates by end of file marker.

OutputFor each test case, output the answer mod 1000000007

Sample Input

1 11 12 11 2

Sample Output

126

Authorhanshuai

Source2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

題意：求不可以操作的區間， 我們用  n-count(可以操作的區間)

 

import java.io.*;import java.math.BigInteger;import java.util.*;public class Main{int n,m,Max=10000010,count=0;BigInteger  mod=BigInteger.valueOf(1000000007);int patten[]=new int[Max];public static void main(String args[]){new Main().work();}void work(){Scanner sc=new Scanner(new BufferedInputStream(System.in));while(sc.hasNext()){n=sc.nextInt();m=sc.nextInt();init();for(int i=0;i<m;i++){int a=sc.nextInt();a--;//給定資料[a,b]我們要求(a,b]這樣無須考慮端點問題int b=sc.nextInt();union(a,b);}System.out.println(pow(26,(n-count)).mod(mod));}}//快速冪取餘BigInteger pow(int a,int b){BigInteger sum=BigInteger.ONE;BigInteger big=BigInteger.valueOf(a);while(b!=0){if((b&1)!=0)sum=(sum.multiply(big)).mod(mod);big=(big.multiply(big)).mod(mod);b>>=1;}return sum;}//並查集合并void union(int a,int b){int pa=fun(a);int pb=fun(b);if(pa==pb)return;if(pa>pb){count++;patten[pb]=pa;}else{count++;//不同的字串，即可以操作的區間，然後用n-count；patten[pa]=pb;}}//並查集尋找int fun(int x){if(patten[x]==x)return x;patten[x]=fun(patten[x]);//路徑壓縮return patten[x];}//初始化void init(){count=0;for(int i=0;i<=n;i++){patten[i]=i;}}}