BZOJ2121 字符串游戏 【dp】

题目链接

BZOJ2121

题解

dp怎么那么神呐QAQ

我们要求出最小字符串长度
我们设一个\(dp[i]\)表示前\(i\)个字符最后所形成的最短字符串长度
对于第\(i\)个字符，要么保留，就是\(dp[i] = dp[i - 1] + 1\)，要么和前面若干个字符一起被删掉

我们设\(c[i][j]\)表示区间\([i,j]\)能否被删掉
如果我们能求出\(c[i][j]\)就好了

我们再设一个\(f[i][j][k][t]\)表示区间\([i,j]\)能否匹配第\(k\)个串的前\(t\)个字符
如果存在一个\(k\)，使得\(f[i][j][k][len[k]]\)为真，那么\(c[i][j]\)就为真

所以我们只需考虑如何求出\(f[i][j][k][t]\)
就是简单的区间\(dp\)啦
枚举断点\(d\)，然后\(f[i][j][k][l] |= f[i][d][k][t] \&\& c[d + 1][j]\)

注意转移的顺序，因为\(f\)和\(c\)是同时计算的，倒序枚举即可

然后就做完啦

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long int
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define BUG(s,n) for (int i = 1; i <= (n); i++) cout<<s[i]<<' '; puts("");
using namespace std;
const int maxn = 155,maxm = 100005,INF = 1000000000;
int f[maxn][maxn][32][23],dp[maxn],c[maxn][maxn],len[32],n,m;
char S[maxn],s[32][23];
int main(){
	scanf("%s",S + 1); m = strlen(S + 1);
	scanf("%d",&n);
	REP(i,n) scanf("%s",s[i] + 1),len[i] = strlen(s[i] + 1);
	for (int i = m; i; i--){
		for (int j = i; j <= m; j++){
			for (int k = 1; k <= n; k++){
				f[i][i - 1][k][0] = 1;
				for (int l = 1; l <= len[k]; l++){
					f[i][j][k][l] = (f[i][j - 1][k][l - 1] && S[j] == s[k][l]);
					for (int d = i; d < j; d++)
						f[i][j][k][l] |= (f[i][d][k][l] && c[d + 1][j]);
				}
			}
			for (int k = 1; k <= n; k++) c[i][j] |= f[i][j][k][len[k]];
		}
	}
	for (int i = 1; i <= m; i++){
		dp[i] = dp[i - 1] + 1;
		for (int j = 1; j <= i; j++) if (c[j][i]) dp[i] = min(dp[i],dp[j - 1]);
	}
	printf("%d\n",dp[m]);
	return 0;
}