UVA

题意：有一排颜色不同的方块，消k个连续方块可得k*k分，问最多得多少分。

分析：

1、solve(l, r, k)---在方块l~r右边再拼上k个颜色等于A[r] 的方块所得到的新序列的最大得分。

2、决策：

(1)直接消去方块r，转移到solve(l, r - 1, 0) + (k + 1)²

(2)枚举i<r,使得a[i]=a[r]，且a[i]!=a[i+1]，转移到solve(i+1,r-1, 0)+solve(l,i,k+1).

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 200 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int dp[MAXN][MAXN][MAXN];
int color[MAXN];
int solve(int l, int r, int k){
    if(l > r) return 0;
    int &ans = dp[l][r][k];
    if(ans) return ans;
    ans = solve(l, r - 1, 0) + (k + 1) * (k + 1);//直接消去方块r
    for(int i = r - 1; i >= l; --i){
        if(color[i] == color[r]){
            ans = max(ans, solve(l, i, k + 1) + solve(i + 1, r - 1, 0));
        }
    }
    return ans;
}
int main(){
    int T;
    scanf("%d", &T);
    int kase = 0;
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i){
            scanf("%d", &color[i]);
        }
        memset(dp, 0, sizeof dp);
        printf("Case %d: %d
", ++kase, solve(1, n, 0));
    }
    return 0;
}