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  • Code Lock[HDU3461]

    Code Lock
    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 2006 Accepted Submission(s): 766


    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

    #include <stdio.h>
    class Union_Find_Set {
    #define MAX_UNION_FIND_SET_SIZE 10000010
    public:
        int setSize;
        int father[MAX_UNION_FIND_SET_SIZE];
        Union_Find_Set() {
            setSize = 0;
        }
        Union_Find_Set(int x) {
            setSize = x;
            clear(x);
        }
        void clear(int x) {
            for (int i = 0; i < x; i++) {
                father[i] = i;
            }
        }
        int getFather(int x) {
            int ret = x, tmp;
            while (ret != father[ret]) {
                ret = father[ret];
            }
            while (x != father[x]) {
                tmp = father[x];
                father[x] = ret;
                x = tmp;
            }
            return ret;
        }
        bool merge(int a, int b) {
            a = getFather(a);
            b = getFather(b);
            if (a != b) {
                father[a] = b;
                return true;
            } else {
                return false;
            }
        }
        int countRoot() {
            int ret = 0;
            for (int i = 0; i < setSize; i++) {
                if (father[i] = i) {
                    ret++;
                }
            }
            return ret;
        }
    };
    Union_Find_Set ufs;
    __int64 quick_Mini(__int64 base, __int64 power, __int64 mod) {
        __int64 x = base, ret = 1;
        while (power) {
            if (power & 1) {
                ret *= x;
                ret %= mod;
            }
            power >>= 1;
            x *= x;
            x %= mod;
        }
        return ret;
    }
    int main() {
        int n, m, a, b, t;
        while (scanf("%d%d", &n, &m) != EOF) {
            ufs.clear(n + 1);
            t = 0;
            for (int i = 0; i < m; i++) {
                scanf("%d%d", &a, &b);
                if (ufs.merge(a - 1, b)) {
                    t++;
                }
            }
            printf("%I64d
    ", quick_Mini(26, n - t, 1000000007));
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/dramstadt/p/6224131.html
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