【BZOJ 2121】字符串游戏

http://www.lydsy.com/JudgeOnline/problem.php?id=2121

dp，设\(f(i,j,k,l)\)表示原串i到j这个子串能否被删成第k个串的长度为l的前缀。

再设\(can(i,j)\)表示原串i到j这个子串能否被删成空串，用can这个状态来加速f的转移即可。

时间复杂度\(O(|L|^3|S||p|)\)，区间dp的转移都很少，所以可以过。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int L = 160;
const int N = 43;
const int p = 33;

bool f[L][L][N][p], can[L][L];
int n, len[N], Slen, dp[L];
char s[N][p], S[L];

int main() {
	scanf("%s", S + 1);
	Slen = strlen(S + 1);
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%s", s[i] + 1);
		len[i] = strlen(s[i] + 1);
	}

	for (int i = Slen; i >= 1; --i)
		for (int j = i; j <= Slen; ++j) {
			for (int k = 1; k <= n; ++k) {
				f[i][i - 1][k][0] = true;
				for (int l = 1; l <= len[k] && l <= j - i + 1; ++l) {
					f[i][j][k][l] |= f[i][j - 1][k][l - 1] && (s[k][l] == S[j]);
					for (int tmp = i; tmp < j; ++tmp) f[i][j][k][l] |= f[i][tmp][k][l] && can[tmp + 1][j];
				}
			}
			for (int k = 1; k <= n; ++k)
				if (f[i][j][k][len[k]]) {
					can[i][j] = true;
					break;
				}
		}

	for (int i = 1; i <= Slen; ++i) {
		dp[i] = dp[i - 1] + 1;
		for (int j = i; j >= 1; --j)
			if (can[j][i])
				dp[i] = min(dp[i], dp[j - 1]);
	}

	printf("%d\n", dp[Slen]);
	return 0;
}