CSU1620: A Cure for the Common Code(KMP+区间DP)

Description

Input

Output

Sample Input

abcbcbcbca
abbbcdcdcdabbbcdcdcd
0

Sample Output

Case 1: 7
Case 2: 11

HINT

Source

题意：把字符串简化，问简化得到的最短长度是多少

思路：要简化首先要求循环节，这里用kmp解决，而要求所有简化中最短的的话，用区间dp可以求得

<pre name="code" class="cpp">#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define LL long long
#define N 505
#define MOD 19999997
#define INF 0x3f3f3f3f
#define EXP 1e-8

int dp[N][N],next1[N],cas = 1;
char s[N];

int kmp(char *s,int len)//kmp求循环节的长度
{
    next1[0] = next1[1] = 0;
    int i;
    UP(i,1,len-1)
    {
        int j = next1[i];
        W((j&&s[i]!=s[j]))
        {
            j = next1[j];
        }
        if(s[i]==s[j])
            next1[i+1]=j+1;
        else
            next1[i+1]=0;
    }
    return len-next1[len];
}

int getbit(int x)
{
    int cnt = 0;
    W(x)
    {
        x/=10;
        cnt++;
    }
    return cnt;
}

void DP()
{
    int n = strlen(s+1);
    int len,l,r,k,i;
    UP(i,1,n)
    dp[i][i]=1;
    UP(len,2,n)//区间dp求解最优值
    {
        for(l = 1; l+len-1<=n; l++)
        {
            r = l+len-1;
            dp[l][r]=dp[l][l]+dp[l+1][r];
            UP(k,l+1,r-1)
            {
                dp[l][r]=min(dp[l][r],dp[l][k]+dp[k+1][r]);
            }
            int t = kmp(s+l,len);
            if(len%t==0)//现在枚举的串是一个周期循环
            {
                int tem = dp[l][l+t-1];
                if(t>1) tem+=2;//因为不是一个字符的话要加括号
                tem+=getbit(len/t);//周期数
                dp[l][r]=min(tem,dp[l][r]);
            }
        }
    }
    printf("Case %d: %d\n",cas++,dp[1][n]);
}

int main()
{
    W(~scanf("%s",s+1))
    {
        if(s[1]=='0')
            break;
        DP();
    }

    return 0;
}