Problem A

Blocks

Input: Standard Input

Output: Standard Output

Time Limit: 10 Seconds

Some of you may have played a game called 'Blocks'. There are n blocks in a row, each box has a color. Here is an example: Gold, Silver, Silver, Silver, Silver, Bronze, Bronze, Bronze, Gold.

The corresponding picture will be as shown below:

Figure 1

If some adjacent boxes are all of the same color, and both the box to its left(if it exists) and its right(if it exists) are of some other color, we call it a 'box segment'. There are 4 box segments. That is: gold, silver, bronze, gold. There are 1, 4, 3, 1 box(es) in the segments respectively.

Every time, you can click a box, then the whole segment containing that box DISAPPEARS. If that segment is composed of k boxes, you will get k*k points. for example, if you click on a silver box, the silver segment disappears, you got 4*4=16 points.

Now let’s look at the picture below:

Figure 2

The first one is OPTIMAL.

Find the highest score you can get, given an initial state ofthis game.

Input

The first line contains the number of tests t(1<=t<=15). Each case contains two lines. The first line contains an integer n(1<=n<=200), the number of boxes. The second line contains n integers, representing the colors of each box. The integers are in the range 1~n.

Output

For each test case, print the case number and the highest possible score.

Sample Input                             Output for Sample Input

Problemsetter: Rujia Liu, Member of Elite Problemsetters' Panel

Special Thanks: Cailiang Liu & Rongjing Xiang from IOI2003 China National Training Team

// 相地，非常有挑性。

1. 先入成的段色 A[], B[] // A : 色, B :

2. 考最左的段要消是不消，如果不消的，另一串接之後消去。

dp[i][j][k] 表示段 [i, j] 左接了 k A[i] 相同色的最大得分。

3. 如果消去最左，得分 = (B[i] + k)^2 + dp[i+1][j][0]

4. 如果不消的，得分 = max(dfs(i+1, pos-1, 0) + dfs(pos, j, k + B[i]))

A[pos] == A[i] // 色相同的那一串接在一起消去。

#include <stdio.h>
#include <algorithm>
using namespace std;
int A[205], B[205];
int dp[205][205][205]; //[l][r][___{l, r}]
char used[205][205][205] = {}, usedcases = 0;
int dfs(int l, int r, int sl) {
    if(l > r)    return 0;
    if(used[l][r][sl] == usedcases)
        return dp[l][r][sl];
    int &ret = dp[l][r][sl];
    ret = 0;
    used[l][r][sl] = usedcases;
    ret = dfs(l+1, r, 0) + (B[l] + sl)*(B[l] + sl);
    for(int i = l+1; i <= r; i++) {
        if(A[i] == A[l]) {
            ret = max(ret, dfs(l+1, i-1, 0) + dfs(i, r, sl + B[l]));
        }
    }
    return ret;
}
int main() {
    int testcase, n, m;
    int i, j, k, cases = 0;
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        for(i = 0; i < n; i++)
            scanf("%d", &A[i]);
        for(i = 1, B[0] = 1, m = 0; i < n; i++) {
            if(A[m] == A[i])
                B[m]++;
            else {
                A[++m] = A[i], B[m] = 1;
            }
        }
        usedcases++;
        printf("Case %d: %d\n", ++cases, dfs(0, m, 0));
    }
    return 0;
}
/*
2
9
1 2 2 2 2 3 3 3 1
1
1
*/