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  • HDU 5842 Lweb and String （水题）

    Lweb and String

    题目链接：

    http://acm.hdu.edu.cn/showproblem.php?pid=5842

    Description

    Lweb has a string S. Oneday, he decided to transform this string to a new sequence. You need help him determine this transformation to get a sequence which has the longest LIS(Strictly Increasing). You need transform every letter in this string to a new number. A is the set of letters of S, B is the set of natural numbers. Every injection f:A→B can be treat as an legal transformation. For example, a String “aabc”, A={a,b,c}, and you can transform it to “1 1 2 3”, and the LIS of the new sequence is 3. Now help Lweb, find the longest LIS which you can obtain from S. LIS: Longest Increasing Subsequence. (https://en.wikipedia.org/wiki/Longest_increasing_subsequence)

    Input

    The first line of the input contains the only integer T,(1≤T≤20). Then T lines follow, the i-th line contains a string S only containing the lowercase letters, the length of S will not exceed 105.

    Output

    For each test case, output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer.

    Sample Input

    2 aabcc acdeaa

    Sample Output

    Case #1: 3 Case #2: 4

    Source

    2016中国大学生程序设计竞赛 - 网络选拔赛
    ##题意： 给出一个字符串，确定一组字符到自然数的一一映射. 使得结果数串的严格LIS最大.
    ##题解： 由于字符串中仅含小写字母，那么结果最大就是26. 所以字符串中有多少个不同的字符，结果就是多少.
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define double long long #define eps 1e-8 #define maxn 101000 #define mod 1000000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;

    char str[maxn];

    int main(int argc, char const *argv[])
    {
    //IN;

    int t, ca=1; cin >> t;
while(t--)
{
    scanf("%s", str);

    int ans = 0;
    int sz = strlen(str);
    int state = 0;
    for(int i=0; i<sz; i++) {
        state |= 1<<(str[i]-'a');
    }

    for(int i=0; i<26; i++) {
        if(state & (1<<i)) ans++;
    }

    printf("Case #%d: %d
", ca++, ans);
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5770779.html
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