POJ 1390 Blocks

最新推荐文章于 2023-04-03 00:11:02 发布

baymax520

最新推荐文章于 2023-04-03 00:11:02 发布

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分类专栏：区间dp 文章标签： dp

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本文介绍了一款名为'Blocks'的游戏，并提供了一种通过动态规划求解游戏最高得分的算法。该算法首先将相同颜色的方块进行合并，然后利用三维动态规划数组dp[i][j][len]来记录区间[i,j]内，右侧还存在长度为len的相同颜色方块时的最大积分。

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Blocks

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5966		Accepted: 2456

Description

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 of this 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

2
9
1 2 2 2 2 3 3 3 1
1
1

Sample Output

Case 1: 29
Case 2: 1

首先将颜色相同的色块合并成大块，用num[i]来表示第i个大块中有多少相同颜色的小块，用c[i]来表示大块的颜色。第一反应想的一般可能是dp[i][j]来表示区间[i,j]的最大积分，那么对于最右边的连续色块有两种操作。第一种就是直接消掉，那么最高分就是dp[i][j-1]+(num[j]*num[j])，第二种就是在j大块之前可能有和j颜色相同的大块k，那么与其合并再消除，最高分为dp[i][k-1]+dp[k+1][j-1]+(num[k]+num[j])*(num[k]+num[j])。

可是这样考虑不全面，因为j与k合并之后不一定是最佳方案，可能在k前还有相同颜色的色块，再和之前的合并才能得到最高分。所以还需要再开一维来表示最右边色块的状态，也就是dp[i][j][len]，表示在j的右边还有长度为len颜色和j一样的色块存在。那么状态转移方程就是dp[i][j][len]=max{dp[i][j-1][0]+(num[j]+len)*(num[j]+len),dp[i][k][len+num[j]]+dp[k+1][j-1][0],(i<=k<=j-2)&&c[k]==c[j]}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=210;
int dp[N][N][N];
int color[N],num[N],c[N];
int f(int i,int j,int len)
{
    if(i==j)
        return dp[i][j][len]=(num[j]+len)*(num[j]+len);
    if(dp[i][j][len])
        return dp[i][j][len];
    int temp=f(i,j-1,0)+(len+num[j])*(len+num[j]);
    for(int k=i;k<j-1;k++)
        if(c[k]==c[j])
        temp=max(temp,f(i,k,num[j]+len)+f(k+1,j-1,0));
    return dp[i][j][len]=temp;
}
int main()
{
    int t,n,kase=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&color[i]);
        int m=0;
        for(int i=1;i<=n;i++)
        {
            if(color[i]==color[i-1])
                num[m]++;
            else
            {
                m++;
                c[m]=color[i];
                num[m]=1;
            }
        }
        printf("Case %d: %d\n",++kase,f(1,m,0));
    }
    return 0;
}