HDU4909 String

String

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

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

3abc?caaabbccaaaaa

 

Sample Output

766
Hint
Good substrings of "abc?ca": "?", "c?", "?c", "c?c", "bc?c", "c?ca", "abc?ca"

题意：给一个由小写字母组成的字符串，里面最多有一个问号‘？’，问有多少个这样的连续字串：

1、字串中的每个字母出现的次数为偶数次

2、？可以代表其中一个字母，也可以删去

题解：本题使用状态压缩 + map + 位运算。

使用状态压缩存储每个位置的状态，需要用到二进制位运算，如果一个字母出现的次数为奇数次，则相应位数为1。由于2^26太大，无法开数组，所以使用map来存储和查找。总的时间复杂度为nlogn

具体见代码：

#include <cstdio>#include <map>#include <cstring>using namespace std;#define N 20005map<int, int> mp1, mp2; map<int, int>::iterator it;char str[N];int ans;void solve(map<int, int> &m1, map<int, int> &m2){for(it = m1.begin(); it != m1.end(); it++){int j = it->first, k = it->second;for(int i = 0; i < 26; i++){//遍历？代表的字母 int q = j ^ (1<<i);if(m2.find(q) != m2.end())ans += k * m2[q];}ans += k * m2[j];//？直接去掉的情况 }}int main(){int t;scanf("%d", &t);while(t--){scanf("%s", str);if(!mp1.empty())mp1.clear();if(!mp2.empty())mp2.clear();int len = strlen(str);int pos = -1;//？的位置 ans = 0;//最终结果 for(int i = 0; i < len; i ++)if(str[i] == '?')pos = i;if(pos != -1) ans++;//只有一个？时，将这个？删掉，也算一种情况 int p = 0;for(int i = pos - 1; i >= 0; i--){p = p ^ (1<<(str[i] - 'a'));//p一直和下一个元素异或 if(p == 0) ans ++;//如果异或为0，那么此时就是偶数，结果+1 ans += mp1[p];//如果之前出现过p，那么现在的p和之前的p异或，结果为0即偶数 mp1[p]++;//p这种情况再+1 }p = 0;for(int i = pos + 1; i < len; i++){p = p ^ (1<<(str[i] - 'a'));if(p == 0) ans ++;ans += mp2[p];mp2[p]++;}if(pos != -1){for(it = mp1.begin(); it != mp1.end(); it++){int j = it->first;int k = it->second;if((j & (j - 1)) == 0)//当只有一位是1时，才可以用？来抵消 ans += k;}for(it = mp2.begin(); it != mp2.end(); it++){int j = it->first;int k = it->second;if((j & (j - 1)) == 0)ans += k;}if(mp1.size() < mp2.size())solve(mp1, mp2);//遍历小的map，搜索大的map else solve(mp2, mp1);}printf("%d\n", ans);}return 0;}