(并查集) Code Lock --HDOJ

Code Lock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 280 Accepted Submission(s): 128

Problem Description

A 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?

Input

There 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.

Output

For each test case, output the answer mod 1000000007

Sample Input

1 1

1 1

2 1

1 2

Sample Output

1

26

Author

hanshuai

Source

2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

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#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<set>
#define mod 1000000007

using namespace std;

int fa[10000005];

int fid(int  x)
{
if(x != fa[x])
{
fa[x] = fid(fa[x]);
}
return fa[x];
}
long long Cal(int a,int n)
{
long long ans = 1,t=a;
while(n)
{
if(n%2)
ans = ans * t % mod;
n /= 2;
t = t * t % mod;
}
return ans;
}
int main(void)
{
//   freopen("in.txt","r",stdin);
int n,m,l,r;
while(scanf("%d %d",&n,&m) != EOF)
{
/*这里的初始化要从零开始，因为我们将L，R转换成了 L-1，R进行
的合并，[1,2] 与 [3,5] 进行合并，等价于[1,5]这个区间
我们需要用到并查集，我们需要把区间调整一下，改成[0,2] 和 [2,5],
也就相当于[0,5] 区间了，这样我们就可以利用并查集进行合并了

*/
for(int i=0; i<=n; ++i)
fa[i] = i;

/*
区间个数最多为n个，即[1,1][2,2][3,3]...[n,n];
每个区间有26个可能，即每个区间都可以是a到z
最后总共有 26^n 种
*/
int cnt = n;

/*
假如我们将[1,1][2,2][3,3]...[n,n]这些区间中前两个合并
就变成了[1,2][3,3]...[n,n];区间少了一个
[1,2] 这个区间有多少种可能那？还是26种
因为aa->bb->cc---->zz 26种
ab->bc->cd---->za 26种
ac->bd->ce---->zb 26种
所以，无论什么组合的两个字符，都是26种可能

每输入一个区间，就判断L-1 与 R是否在一个集合，如果不在，就合并区间
，区间数就少1
*/
for(int i=0; i<m; ++i)
{
scanf("%d %d",&l,&r);
int x = fid(l-1);
int y = fid(r);
if(x != y)
{
fa[x] = y;
cnt--;
//             cout <<cnt<<endl;
}
}
/*
最后计算26的区间个数次方就ＯＫ了
*/
cout<<Cal(26,cnt)<<endl;
}
return 0;
}