【题解】LuoGu2470：[SCOI2007]压缩

原题传送门
这道题跟这道很像，差别在于本题无法嵌套
令$d{p}_{i,j,0/1}$表示区间$[l,r]$中有/无$M$的答案，默认在$i-1$的地方有一个$M$
转移：

• 若区间前半段和后半段相等，$d{p}_{i,j,0}=d{p}_{i,mid,0}+1$
• $d{p}_{i,j,0}=min(d{p}_{i,k}+j-k)$
• $d{p}_{i,j,1}=min(min(d{p}_{i,k,0},d{p}_{i,k,1})+1+min(d{p}_{k+1,j,0},d{p}_{k+1,j,1}))$（枚举$M$的位置）

$$ans=min(d{p}_{1,n,0},d{p}_{1,n,1})$$

Code：

#include 
#define maxn 110
using namespace std;
char s[maxn];
int dp[maxn][maxn][2], n;

bool check(int l, int r){
	if ((r - l + 1) % 2 != 0) return 0;
	int mid = (l + r) >> 1;
	for (int i = l; i <= mid; ++i) if (s[i] != s[mid + i - l + 1]) return 0;
	return 1;
}

int main(){
	scanf("%s", s + 1); n = strlen(s + 1);
	memset(dp, 0x3f, sizeof(dp));
	for (int i = 1; i <= n; ++i)
		for (int j = i; j <= n; ++j) dp[i][j][0] = dp[i][j][1] = j - i + 1;
	for (int l = 2; l <= n; ++l)
		for (int i = 1, j = i + l - 1; j <= n; ++i, ++j){
			if (check(i, j)) dp[i][j][0] = dp[i][(i + j) >> 1][0] + 1;
			for (int k = i; k < j; ++k) dp[i][j][0] = min(dp[i][j][0], dp[i][k][0] + j - k);
			for (int k = i; k < j; ++k) dp[i][j][1] = min(dp[i][j][1], min(dp[i][k][0], dp[i][k][1]) + min(dp[k + 1][j][0], dp[k + 1][j][1]) + 1);
		}
	printf("%d\n", min(dp[1][n][0], dp[1][n][1]));
	return 0;
}