UVA 11552 Fewest Flops （序列划分DP，4级）

D - Fewest Flops
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Time Limit:2000MS
Memory Limit:0KB 64bit IO Format:%lld & %llu

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Status

Appoint description:
System Crawler (2013-05-30)

Description




Problem F

FEWEST FLOPS

A common way to uniquely encode a string is by replacing its consecutive repeating characters (or “chunks”) by the number of times the character
occurs followed by the character itself. For example, the string “aabbbaabaaaa” may be encoded as “2a3b2a1b4a”. (Note for this problem even a single character “b” is replaced by “1b”.)

Suppose we have a string
S and a number k such that k divides the length of
S. Let S1 be the substring of
S from 1 to k, S2 be the substring of
S from k + 1 to 2k, and so on. We wish to rearrange the characters of each block
Si independently so that the concatenation of those permutations
S’ has as few chunks of the same character as possible. Output the fewest number of chunks.

For example, let
S be “uuvuwwuv” and k be 4. Then S1 is “uuvu” and has three chunks, but may be rearranged to “uuuv” which has two chunks. Similarly,
S2 may be rearranged to “vuww”. Then
S’, or S1S2, is “uuuvvuww” which is 4 chunks, indeed the minimum number of chunks.

Program Input

The input begins with a line containing
t (1 ≤ t ≤ 100), the number of test cases. The following t
lines contain an integer k and a string S made of no more than 1000 lowercase English alphabet letters. It is guaranteed that
k will divide the length of S.

Program Output

For each test case, output a single line containing the minimum number of chunks after we rearrange
S as described above.

INPUT

2
5 helloworld
7 thefewestflops

OUTPUT

8
10

Calgary Collegiate Programming Contest 2008

序列划分
思路：当前块有几个，最后一个是哪个

#include<iostream>
#include<cstring>
#include<cstdio>
#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=1e3+9;
const int oo=0x3f3f3f;
int dp[mm][27],has[mm][27],pos;
char s[mm];
int main()
{ int cas,K;
while(~scanf("%d",&cas))
{
while(cas--)
{ pos=0;
scanf("%d%s",&K,s);
int len=strlen(s);
clr(has,0);clr(dp,0x3f);
FOR(i,1,len)
{
if(i%K==1||K==1)++pos;
++has[pos][s[i-1]-'a'];
}
FOR(i,0,25)dp[0][i]=1;
int num;
FOR(i,1,pos)
{ FOR(j,0,25)
{int&ret=dp[i][j];
num=0;
if(has[i][j]==0)continue;
FOR(k,0,25)
if(has[i][k])++num;
FOR(k,0,25)
if(has[i][k]&&k!=j)
ret=min(ret,num-1+dp[i-1][k]);
else if(num==1&&k==j)ret=min(ret,dp[i-1][k]);
else ret=min(ret,num+dp[i-1][k]);
}
}
int ans=oo;
FOR(i,0,25)ans=min(ans,dp[pos][i]);
printf("%d\n",ans);
}
}
return 0;
}