Abbreviating each color to a single uppercase letter, Karel would
write the stripe pattern red-green-blue-green-red as "RGBGR"
(quotes added for clarity).  It would take him three brush strokes to
paint this pattern.
The first stroke would cover the entire canvas in red (RRRRR).
The second stroke would leave a band of red on either side and fill
in the rest with green (RGGGR).
The final brush stroke would fill in the blue stripe in the center
(RGBGR).

Given a stripe pattern stripes as a String, calculate the minimum number of
brush strokes required to paint that pattern.

DEFINITION
Class:StripePainter
Method:minStrokes
Parameters:String
Returns:int
Method signature:int minStrokes(String stripes)

NOTES
-The blank canvas is an ugly color and must not show through.

CONSTRAINTS
-stripes will contain only uppercase letters ('A'-'Z', inclusive).
-stripes will contain between 1 and 50 characters, inclusive.

EXAMPLES

0)
"RGBGR"

Returns: 3

Example from introduction.

1)
"RGRG"

Returns: 3

This example cannot be done in two strokes, even though there are only two colors.
Suppose you tried to paint both red stripes in one stroke, followed by both green stripes
in one stroke.  Then the green stroke would cover up the second red stripe.  If you tried
to paint both green stripes first, followed the red stripes, then the red stroke would
cover up the first green stripe.

2)
"ABACADA"

Returns: 4

One long stroke in color 'A', followed by one stroke each in colors 'B', 'C', and 'D'.

3)
"AABBCCDD CCBB AABBCCDD"

Returns: 7

4)
"BECBBDDEEBABDCADEAAEABCACBDBEECDEDEACACCBEDABEDADD"

Returns: 26

class StripePainter {

    static int[][] dp;//dp[i][j] 表示从i-j的范围内 最小的涂色次数
    static int n;

    public int minStrokes(String s) {
        int res = 0;
        dp = new int[s.length()][s.length()];
        n = s.length();

        for (int i = 0; i < n; i++) {
            dp[i][i] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 0, t = j + i; j < n && t < n; j++, t = j + i) {
                int min = Integer.MAX_VALUE;
                for (int k = j; k < t; k++) {
                    int tmp = dp[j][k] + dp[k + 1][t];
                    if (s.charAt(j) == s.charAt(t))
                        tmp -= 1;
                    if (min > tmp)
                        min = tmp;
                }
                dp[j][t] = min;
            }
        }

        return dp[0][s.length() - 1];
    }
}