poj1390 2010.8.2

Poj 1390  Blocks

 

【关键字】

DP 记忆化搜索

【摘要】

给你一行方块，每个方块都有自己的颜色，连续的同色方块可以消除，得到的分数是个数x^2，求消除所有方块后的得分最大值。

【正文】

1、题目描述

Blocks

Time Limit: 5000MS Memory Limit: 65536K

Total Submissions: 1916  Accepted: 734

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 anexample: Gold, Silver, Silver, Silver, Silver, Bronze, Bronze, Bronze, Gold.

The corresponding picture will be as shownbelow:

 

Figure 1

If some adjacent boxes are all of the samecolor, 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 thesegments respectively.

 

Every time, you can click a box, then thewhole segment containing that box DISAPPEARS. If that segment is composed of kboxes, you will get k*k points. for example, if you click on a silver box, thesilver 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, givenan initial state of this game.

Input

The first line contains the number of testst(1<=t<=15). Each case contains two lines. The first line contains aninteger n(1<=n<=200), the number of boxes. The second line contains nintegers, representing the colors of each box. The integers are in the range1~n.

Output

For each test case, print the case numberand the highest possible score.

SampleInput

2

9

1 2 2 2 2 3 3 3 1

1

1

SampleOutput

Case 1: 29

Case 2: 1

Source

Source

lrj黑书

 

2、算法分析

根据小方块的规则，连在一起的相同颜色的方块可以合并。color[i]表示第i段颜色，len[i]表示第i段的方块的长度。

f[I,j,k]表示把（color[i],len[i]）,(color[i+1],len[i+1]),……(color[j-1],len[j-1]),(color[j],color[j]+k)合并的得分。

考虑(color[j],len[j]+k)这段，要么马上消掉，要么和前面若干段一起消掉

1．马上消掉 f[I,j-1,0]+(len[j]+k)^2;

2．若干段，若干段最后面的那一段是p,  f[I,p,k+len[j]]+f[p+1,j-1,0];

f[I,j,k]=max{ f[I,j-1,0]+(len[j]+k)^2, f[I,p,k+len[j]]+f[p+1,j-1,0]}  (color[p]==color[j])

边界  f[I,i-1,0]=0

3、源码

#include <cstdio>#include <cstring>#define MAXN 210#define INF -1int color[MAXN],len[MAXN];int n,num;int f[MAXN][MAXN][MAXN];void init(){ memset(f,-1,sizeof(f)); scanf("%d",&n); int temp; scanf("%d",&temp); color[1]=temp; len[1]=1; num=1; for(int i=2;i<=n;i++) {  scanf("%d",&temp);  if (temp==color[num])   len[num]++;  else  {   num++;   color[num]=temp;   len[num]=1;  } }}int dpit(int i,int j,int k){ if (f[i][j][k]!=-1)  return f[i][j][k]; if (i==j+1) return 0; int ans=INF; int temp=dpit(i,j-1,0)+(len[j]+k)*(len[j]+k); if (temp>ans)  ans=temp; for(int p=i;p<j;p++)  if (color[p]==color[j])  {   temp=dpit(i,p,len[j]+k)+dpit(p+1,j-1,0);   if (temp>ans)    ans=temp;  } f[i][j][k]=ans; return ans;}int main(){ int t; scanf("%d",&t); for (int i=1;i<=t;i++) {  init();  int ans=dpit(1,num,0);  printf("Case %d: %d\n",i,ans); } return 0;}