CODEFORCES 264D Colorful Stones

题目：http://codeforces.com/problemset/problem/264/D

题意:R、B、G组成的两个排石头组，松鼠和猫分别从石头的左端到右端，某一个时刻，你可以允许一种颜色的石头的石头通行。问松鼠和猫可以有多少组相对关系？

分析：状态、字符串，计数

Input

RBR
RGG

Output

5

Input

RGBB
BRRBRR

Output

19

Input

RRRRRRRRRR
RRRRRRRR

Output

8

Note

In the first example, there are five reachable states: (1, 1), (2, 2), (2, 3), (3, 2), and (3, 3). For example, the state (3, 3) is reachable because if you perform instructions "RED", "GREEN", and "BLUE" in this order from the initial state, the state will be (3, 3). The following picture shows how the instructions work in this case.

#include 
#include 
#include 
#include 
#define change(x) (x == 'R' ? 0 : (x == 'G' ? 1 : 2))
#define LL long long

using namespace std;

void tension(int &a, int &b)
{
    if (b < a)
        a = b;
}

const int MAXN =(int) 1e6 + 3;

int n, m;
char s[MAXN], t[MAXN];
int p1[MAXN], p2[MAXN];
int sum[MAXN][3][3];

void doing(char* const s, int n, char* const t, int m, int* const num)
{
    for (int i = 1; i <= n; i ++)
    {
        num[i] = num[i - 1] + 1;
        while (num[i] < m && t[num[i]] != s[i])
            num[i] ++;
        tension(num[i], m);
    }
}

int main()
{
    //printf("Hello world!\n");
    scanf("%s\n%s", s + 1, t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    doing(s, n, t, m, p1);
    doing(t, m, s, n, p2);
    for (int i = 2; i <= m; i ++)
    {
        for (int j = 0; j < 3; j ++)
            for (int k = 0; k < 3; k ++)
            {
                if (change(t[i - 1]) == j && change(t[i]) == k)
                    sum[i][j][k] = sum[i - 1][j][k] + 1;
                else
                    sum[i][j][k] = sum[i - 1][j][k];
            }
    }

    LL ans = 0;
    int L = 1;
    for (int i = 1; i <= n; i ++)
    {
        int R = p1[i];
        while (L <= m && p2[L] < i) L ++;
        ans += max(R - L + 1, 0);
        // printf("(%d,%d\n",i,L);
        if (L <= R && i > 1)
        {
            int j = change(s[i - 1]);
            int k = change(s[i]);
            if (j != k)
            {
                ans -= sum[R][k][j];
                ans += sum[L - 1][k][j];
            }
        }
    }
    cout << ans << endl;
    return 0;
}