poj 1390 Blocks （经典区间dp 方块消除）

Blocks

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4250		Accepted: 1704

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

Source

Liu Rujia@POJ

题意：

一排有颜色的方块，每次能够消除相邻的颜色同样的块，得分为方块个数的平方。消除后剩下的方块会合并，问如何消除方块使得总得分最大。

思路：

黑书原题（p123），合并初始相邻同样的块，得到颜色数组c和相应的长度len，dp[i][j][k]表示i~j区间，与后面k个同样颜色块一起消除得分的最大值（当然k个块的颜色必须与j同样），考虑len[j]和k这一段怎么消除，有两种可能：

1.单独消除，dp[i][j][k]=dp[i][j-1][0]+(len[j]+k)^2;

2.和前面的一起消除，如果前面的一起消除的块最后一块为p，那么dp[i][j][k]=dp[i][p][k+len[j]]+dp[p+1][j-1][0]。

能够依据p和j的颜色同样以及k的范围来优化一下。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 205
#define MAXN 200005
#define INF 0x3f3f3f3f
#define mod 1000000007
#define eps 1e-6
const double pi=acos(-1.0);
typedef long long ll;
using namespace std;

int n,m,ans,tot;
int a[maxn],c[maxn],len[maxn],pos[maxn],last[maxn];
int dp[205][205][205],num[maxn][maxn];

void solve()
{
    int i,j,k,p;
    memset(pos,0,sizeof(pos));
    for(i=1;i<=tot;i++)
    {
        last[i]=pos[c[i]];
        pos[c[i]]=i;
    }
    memset(num,0,sizeof(num));
    for(i=tot;i>=1;i--)
    {
        for(j=1;j<=n;j++)
        {
            if(j==c[i]) num[j][i]=num[j][i+1]+len[i];
            else num[j][i]=num[j][i+1];
        }
    }
    memset(dp,0,sizeof(dp));
    for(int l=1;l<=tot;l++)
    {
        for(i=1;i<=tot;i++)
        {
            j=i+l-1;
            if(j>tot) break ;
            for(k=0;k<=num[c[j]][j+1];k++)
            {
               dp[i][j][k]=dp[i][j-1][0]+(len[j]+k)*(len[j]+k);
               for(p=last[j];p>=i;p=last[p])
               {
                   dp[i][j][k]=max(dp[i][j][k],dp[i][p][len[j]+k]+dp[p+1][j-1][0]);
               }
            }
        }
    }
    ans=dp[1][tot][0];
}
int main()
{
    int i,j,test,ca=0;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        tot=0;
        memset(len,0,sizeof(len));
        for(i=1;i<=n;)
        {
            tot++;
            c[tot]=a[i];
            while(i<=n&&a[i]==c[tot]) i++,len[tot]++;
        }
        solve();
        printf("Case %d: %d\n",++ca,ans);
    }
    return 0;
}