bestcoder3(1003)hdu4909(状态压缩+乱搞)

最新推荐文章于 2016-07-06 22:48:05 发布

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最新推荐文章于 2016-07-06 22:48:05 发布

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分类专栏： bestcoder 文章标签： ACM hdu bestcoder

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本文探讨了在给定字符串中包含至多一个问号的情况下，如何找出所有满足条件的子串，即每个子串中字母出现次数为偶数次。通过二进制压缩状态的方法，简化了状态转换和子串匹配的过程。

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String

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 443 Accepted Submission(s): 113

Problem Description

You hava a non-empty string which consists of lowercase English letters and may contain at most one '?'. Let's choose non-empty substring G from S (it can be G = S). A substring of a string is a continuous subsequence of the string. if G contains '?' then '?' can be deleted or replaced by one of lowercase english letters. After that if each letter occurs even number of times in G then G is a good substring. Find number of all good substrings.

Input

The input consists of an integer T, followed by T lines, each containing a non-empty string. The length of the string doesn't exceed 20000.

[Technical Specification]
1 <= T <= 100

Output

For each test case, print a single integer which is the number of good substrings of a given string.

Sample Input

3
abc?ca
aabbcc
aaaaa

Sample Output

7
6
6
Hint
Good substrings of "abc?ca": "?", "c?", "?c", "c?c", "bc?c", "c?ca", "abc?ca"

题意：就是给一个字符串，该字符串最多含一个问号，然后问你有多少个满足条件的子串，满足条件定义为，该子串里面的所有字符都出现偶数次，如果含有问号，可以将问号替换为任意字符也可以不替换（如果不替换，问号就算作空字符）

思路：由于只有26个小写字母，可以用26位的二进制来压缩子串的状态，那么首先对整个字符串处理，for一遍以后，每个位置都有一个对应的值（将该位置的字符异或到对应的位置上），如果有两个位置对应的值相同，那么这两个位置之间的子串就是满足条件的，可以分两种情况

1.不含问号，那么去掉问号处理一遍即可

2.含问号，那么可以枚举问号为26个字符，再分别处理每一次的情况即可

这题有点卡时，当时被坑了好久~

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define INF ((1<<31)-1)

int n;
char s[20010];
int ans[20010];
int ha[20010];
int jj[27];
int ii[27];

int A[20010];
int B[20010];

int main()
{
    int t,i,j;

    scanf("%d",&t);

    while(t--)
    {
        memset(s,0,sizeof(s));
        scanf("%s",s);
        n=strlen(s);

        int flag=-1;

        for(i=0;i>1;

        if(flag==-1)printf("%d\n",ans1);

        else {

            d=0;
            temp=0;
            ans[d++]=0;
            for(i=0;iB[i2])i2++;
                else {

                    int cnt1=1;
                    int cnt2=1;
                    while(i1