Hdu 6212 Zuma【思维+区间Dp】

Zuma

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 363    Accepted Submission(s): 103

Problem Description

Think about the Zuma Game. You have a row of at most 200 black(0) or white(1) balls on the table at the start. Each three consecutive balls never share the same colour. You also have infinite amount of black and white balls in your hand. On each turn, you can choose a ball in your hand and insert it into the row, including the leftmost place and the rightmost place. Then, if there is a group of three of more balls in the same colour touching, remove these balls. Keep doing this until no more balls can be removed.
Find the minimal balls you have to insert to remove all the balls on the table.

 

Input

The first line of input contains an integer T (1≤T≤100) which is the total number of test cases.
Each test case contains a line with a non-empty string of 0 and 1 describing the row of balls at the start.

 

Output

For each test case, output the case number and the minimal balls required to insert in a line.

 

Sample Input

4
10101
101001001
1001001001
01001101011001100

 

Sample Output

Case #1: 4
Case #2: 3
Case #3: 3
Case #4: 2

题目大意：

一共有两种颜色的祖玛游戏，每三个连在一起（或者大于三个）的球球就会被消除掉，问将这个字符串消除干净的最小吐球个数。

思路（思路源自：http://blog.youkuaiyun.com/zchahaha/article/details/78026747）：

2006年百度之星复赛，zuma。原题。

①我们观察到数据范围不大，又要搞定最优解问题，我们不妨考虑区间Dp，设定Dp【i】【j】表示消除区间从i到j之间的球球的最小吐球个数。

同时将字符串处理到a【】，a【i】表示第i个连续相同颜色的球球的个数。

②那么我们不难想到几种消除方式：

问题肯定是从小区间递推到大区间的。

1.将区间分成两部分，各消除自身的部分：Dp【i】【j】=min（Dp【i】【j】，Dp【i】【k】+Dp【k+1】【j】）；

2.将中间区间全部消除掉之后，两端的相同颜色相撞删除：Dp【i】【j】=min（Dp【i】【j】，Dp【i+1】【j-1】+（a【i】+a【j】是否小于3））；

3.中间有1个球球，然后消除掉其左右两侧的部分之后，3部分相撞消除：Dp【i】【j】=min（Dp【i】【j】，Dp【i+1】【k-1】+Dp【k+1】【j-1】）；此时要求a【i】+a【j】<4，因为a【i】和a【j】都至少是1，如果相加大于等于4的话，无论先删除左侧区间还是右侧区间，都会构成2消的情况，而达不成三消的情况。

过程维护一下最优解即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char s[250];
int a[250];
int dp[250][250];
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",s);
        int n=strlen(s);
        int m=1;
        a[1]=1;
        for(int i=1;i<n;i++)
        {
            if(s[i]==s[i-1])a[m]++;
            else a[++m]=1;
        }
        for(int len=0;len<=m;len++)
        {
            for(int i=1;i<=m;i++)
            {
                int j=i+len;
                if(j<=m&&j>=1)
                {
                    dp[i][j]=2*n;
                    if(len==0)
                    {
                        dp[i][j]=3-a[i];
                    }
                    else
                    {
                        for(int k=i;k<j;k++)
                        {
                            dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
                        }
                        if((j-i-1)%2==1)
                        {
                            if(a[i]+a[j]==2)
                            dp[i][j]=min(dp[i][j],dp[i+1][j-1]+1);
                            else dp[i][j]=min(dp[i][j],dp[i+1][j-1]);
                            if(a[i]+a[j]<4)
                            for(int k=i+2;k<j;k+=2)
                            {
                                if(a[k]==1)
                                dp[i][j]=min(dp[i][j],dp[i+1][k-1]+dp[k+1][j-1]);
                            }
                        }
                    }
                }
            }
        }
        printf("Case #%d: ",++kase);
        printf("%d\n",dp[1][m]);
    }
}