【bzoj1090】[SCOI2003]字符串折叠

1090: [SCOI2003]字符串折叠

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1652  Solved: 1093

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Description

折叠的定义如下： 1. 一个字符串可以看成它自身的折叠。记作S  S 2. X(S)是X(X>1)个S连接在一起的串的折叠。记作X(S)  SSSS…S(X个S)。 3. 如果A  A’, BB’，则AB  A’B’ 例如，因为3(A) = AAA, 2(B) = BB，所以3(A)C2(B)  AAACBB，而2(3(A)C)2(B)AAACAAACBB
给一个字符串，求它的最短折叠。例如AAAAAAAAAABABABCCD的最短折叠为：9(A)3(AB)CCD。

Input

仅一行，即字符串S，长度保证不超过100。

Output

仅一行，即最短的折叠长度。

Sample Input

NEERCYESYESYESNEERCYESYESYES

Sample Output

14

HINT

一个最短的折叠为：2(NEERC3(YES))

Source

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区间DP

设f[i][j]为区间[i,j]的答案

显然有f[i][j] = min{f[i][k] + f[k + 1][j]} (i <= k < j)
如果[i,k]能够折叠[k + 1,j]的时候，还有f[i][j]  = min{f[i][k] + 2 + bit((j - i + 1) / (k - i + 1))} 

其中bit(x)为计算x的十进制位数的函数

代码：

#include<cstdio>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;

typedef long long LL;

const int INF = 2147483647;
const int maxn = 110;

char s[maxn];
int n,f[maxn][maxn],maxlen[maxn][maxn];

inline LL getint()
{
LL ret = 0,f = 1;
char c = getchar();
while (c < '0' || c > '9')
{
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9')
ret = ret * 10 + c - '0',c = getchar();
return ret * f;
}

inline int cal(int x)
{
int cnt = 0;
while (x) cnt++ , x /= 10;
return cnt;
}

int main()
{
scanf("%s",s + 1);
n = strlen(s + 1);
for (int i = 1; i <= n; i++)
for (int j = i; j <= n; j++)
for (int k = 1; k <= n; k++)
{
bool tmp = 0;
if (i + k * (j - i + 1) > n) break;
for (int l = 0; l <= j - i; l++)
if (s[i + l] != s[i + k * (j - i + 1) + l])
{
tmp = 1;
break;
}
if (tmp) break;
maxlen[i][j] = k;
}
for (int i = 1; i <= n; i++) f[i][i] = 1;
for (int len = 1; len <= n; len++)
for (int i = 1; i <= n; i++)
{
int j = i + len;
int test;
if (i == 6 && j == 14)
test = 1;
f[i][j] = INF;
for (int k = i; k < j; k++)
f[i][j] = min(f[i][j],f[i][k] + f[k + 1][j]);
for (int k = i; k < j; k++)
{
if ((j - i + 1) % (k - i + 1) != 0) continue;
if (maxlen[i][k] + 1 >= (j - i + 1) / (k - i + 1))
f[i][j] = min(f[i][j],f[i][k] + 2 + cal((j - i + 1) / (k - i + 1)));
}
}
printf("%d",f[1]
);
return 0;
}