Code Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1308    Accepted Submission(s): 471

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

思想很好。。。代码调不粗来。。真是渣啊。。。

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include<queue>
using namespace std;
#define INF 99999999
const int maxn =600;
const int maxm =360000;
const int oo = 1<<29;
int f[11000000];
#define mod 1000000007
#define LL __int64
int find(int x)
{
    while(x!=f[x])x=f[x];
    return x;
}
LL mul(LL a,LL b)
{
    if(b==0)return 1;
    if(b==1)return a;
    LL c=mul(a,b/2);
    c=(c*c)%mod;
    if(b&1)c=(c*a)%mod;
    return c;
}
int main()
{
    int n,m,a,b,i;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<=n;i++)f[i]=i;
        LL ans=n;
        for(i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            a=find(a-1);
            b=find(b);
            if(a==b)continue;
            f[b]=a;
            ans--;
        }
        cout<<mul(26,ans)<<endl;
    }
    return 0;
}