BZOJ 1090 [SCOI2003]字符串折叠

题目链接

用KMP判断最小循环节

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
char s[105];
int f[105][105];
int fail[105];
int calc(int l, int r) {
	fail[l] = fail[l + 1] = l;
	for(int i = l + 1; i <= r; i++) {
		int cur = fail[i];
		while(cur != l && s[cur] != s[i]) cur = fail[cur];
		if(s[cur] == s[i]) cur++;
		fail[i + 1] = cur;
	}
	int len1 = r - l + 1, len2 = len1 - (fail[r + 1] - l);
	int res = len1 + 3;
	if(len1 % len2 == 0) {
		res = f[l][len2 + l - 1] + 3;
		if(len1 / len2 > 9) res++;
		if(len1 / len2 > 99) res++;
	}
	return res;
}
int main() {
	scanf("%s", s);
	int n = strlen(s);
	memset(f, 0x3f, sizeof(f));
	for(int i = 0; i < n; i++) f[i][i] = 1;
	for(int len = 1; len <= n; len++) {
		for(int l = 0, r = len - 1; r < n; l++, r++) {
			for(int i = l; i < r; i++) 
			  f[l][r] = min(f[l][r], f[l][i] + f[i + 1][r]);
			f[l][r] = min(f[l][r], calc(l, r));
		}
	}
	printf("%d\n", f[0][n - 1]);
	return 0;
}