UVALive 3363 String Compression （区间DP，4级）

最新推荐文章于 2018-11-08 22:11:14 发布

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本文介绍了一种基于重复子串的字符串压缩方法，通过寻找并替换重复出现的子串来减少存储空间。讨论了压缩规则，并提供了一个实现该压缩算法的程序示例。

M - String Compression

Crawling in process... Crawling failed Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu

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Appoint description:

Description

Run Length Encoding(RLE) is a simple form of compression. RLE consists of the process for searching for a repeated runs of a single character in a string to be compressed, and replacing them by a single instance of the character and a run count. For example, a string abcccddddddefgggggggggghijk is encoded into a string ab3c6def10ghijk by RLE.

A new compression method similar to RLE is devised and the rule of the method is as follows: if a substring S is repeated k times, replace k copies of S by k(S) . For example, letsgogogo is compressed into lets3(go). The length of letsgogogo is 10, and the length of lets3(go) is 9. In general, the length of k(S) is (number of digits in k ) + (length of S ) + 2 (for `(' and `)'). For example, the length of 123(abc) is 8. It is also possible to nest compression, so the substring S may itself be a compressed string. For example, nowletsgogogoletsgogogo could be compressed as a now2(lets3(go)), and nowletsgogogoletsgogogoandrunrunrun could be compressed as now2(lets3(go))and3(run).

Write a program that, for a given string, gives a shortest compressed string using the compression rules as described above.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case consists of a single line containing one string of no more than 200 characters drawn from a lower case alphabet. The length of shortest input string is 1.

Output

Your program is to write to standard output. Print exactly one line for each test case. For each test case, print the length of the shortest compressed string.

The following shows sample input and output for four test cases.

Sample Input

4 
ababcd 
letsgogogo 
nowletsgogogoletsgogogo 
nowletsgogogoletsgogogoandrunrunrun

Sample Output

字符串压缩。

思路：区间DP，划分区间而后合并，dp[i][j] 区间i-j的最优值。

     dp[i][j]=min(dp[i][k]+dp[k+1][j],i-j区间合并）

#include<iostream>
#include<cstdio>
#include<cstring>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=201;
int dp[mm][mm];
int bit[mm];
char s[mm];
bool ok(int l,int r,int dis)
{
  FOR(i,l,l+dis-1)
  for(int j=i+dis;j<=r;j+=dis)
    if(s[i]!=s[j])return 0;
  return 1;
}
void DP(int l,int r)
{
  if(l==r){dp[l][r]=1;return;}
  int&ret=dp[l][r];
  ret=mm;
  FOR(i,l,r-1)ret=min(ret,dp[l][i]+dp[i+1][r]);
  int len=r-l+1;
  //if(len&1)return;3的倍数也行 O(∩_∩)O~
  FOR(dis,1,len)
  { 
    if(len%dis==0&&ok(l,r,dis))//整数倍
      ret=min(ret,dp[l][l+dis-1]+2+bit[len/dis]);
  }
}
void BIT()
{
  FOR(i,0,9)bit[i]=1;
  FOR(i,10,99)bit[i]=2;
  FOR(i,100,200)bit[i]=3;
}
int main()
{
  int cas;BIT();
  while(~scanf("%d",&cas))
  {
    while(cas--)
    {
      scanf("%s",s);
      int len=strlen(s);
      clr(dp,-1);
      FOR(dis,1,len)
      for(int i=0;i+dis<=len;++i)
      DP(i,i+dis-1);
      printf("%d\n",dp[0][len-1]);
    }
  }
}