H.High String

原题：

Description:

We define the srting which can satisfy the following conditions as High String:

1)The string only contains six kinds of characters: A,B,C,D,E and F;

2)cont(A) = count(b), count(c) = cont(D),and count(E) = count(F). count(X) means the number of character X in the string;

3) Except the string itself, there won't be any substrings can satisfy the second condition.

1<=length(S)<=100000

 

Sample Input:

ACBEDF

ACBD

GABCD

ACBBD

Sample Output:

YES

YES

NO

NO

类型：

字符串hash

题意：

判断字符串是否满足下面三个条件：

1)只包含ABCDEF六个字符

2)A的数量=B的数量，C的数量=D的数量，E的数量=F的数量

3)除了自身，任何子串不满足条件二

思路：

设i-j子串能满足条件二，则有

count[j]['A'] - count[i]['A'] = count[j]['B'] - count[i]['B'];

交换得

count[j]['A'] - count[j]['B'] = count[i]['A'] -count[i]['B'];

这样，只要求出之前的AB，CD，EF数量差，然后用hash查找当前的差是否出现过，这样复杂度就为O(n);

附代码（未AC验证）：

#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;

struct node {
    int ab,cd,ef;
};
vector <node> p[1000003];
char str[100100];


int hash(int a, int b, int c) {
    int seed = 13131;
    int ans = 1;
    ans *= a;
    ans *= seed;
    ans *= b;
    ans *= seed;
    ans *= c;
    return ans%1000003;
}

int main() {
    while (scanf("%s", str) != EOF) {
        int i;
        int len = strlen(str);
        int num[6] = {0};
        int succeed = 1;

        for (i = 0; i <= 1000003; i++) {
            p[i].clear();
        }
        
        for (i = 0; i < len && succeed; i++) {
            if (str[i] > 'F') {
                succeed = 0;
                break;
            }
            num[str[i]-'A']++;

            int ab = num[0]-num[1];
            int cd = num[2]-num[3];
            int ef = num[4]-num[5];

            int hashn = hash(ab, cd, ef);
            int size = p[hashn].size();
            int j;
            for (j = 0; j < size; j++) {
                node &pt = p[hashn][j];
                if (pt.ab == ab && pt.cd == cd && pt.ef == ef) {
                    succeed = 0;
                    break;
                }
            }

            node t;
            t.ab=ab, t.cd=cd, t.ef=ef;
            p[hashn].push_back(t);
            //printf("p[%d].size() = %d\n", hashn, p[hashn].size());
        }
        
        if (num[0] != num[1] || num[2] != num[3] || num[4] != num[5]) {
            succeed = 0;
        }

        if (!succeed) puts("NO");
        else puts("YES");
    }
    return 0;
}