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(并查集) hdu 3461

2024-11-21 13:08:40   202人阅读

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 11 12 11 2
 

 

Sample Output
126
 

 

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 99999999const int maxn =600;const int maxm =360000;const int oo = 1<<29;int f[11000000];#define mod 1000000007#define LL __int64int 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;}

　　

(并查集) hdu 3461


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