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  • BZOJ 5336: [TJOI2018]party

    状压最长公共子序列的DP数组,一维最多K(15)个数,且相邻两个数的差不超过1,2^15种状态,预处理转移

    #include<cstdio>
    #include<algorithm>
    using namespace std;
    const int mod=1e9+7;
    int n,K,now[25],G[25],To[50005][3],F[2][50005][3],ED[50005],ANS[25],S[25];
    char s[1000005];
    void dfs(int t,int s){
    	if (t>K){
    		for (int i=1; i<=K; i++) {
    			now[i]=now[i-1];
    			if (s&(1<<i-1)) now[i]++;
    		}
    		ED[s]=now[K];
    		for (int to=0; to<3; to++){
    			for (int i=1; i<=K; i++) G[i]=max(max(G[i-1],now[i]),now[i-1]+(to==S[i]));
    			for (int i=K; i>=1; i--) (To[s][to]<<=1)|=(G[i]-G[i-1]);
    		}
    		return;
    	}
    	dfs(t+1,s<<1);
    	dfs(t+1,s<<1|1);
    }
    int main(){
    	scanf("%d%d",&n,&K);
    	scanf("%s",s+1);
    	for (int i=1; i<=K; i++){
    		if (s[i]=='N') S[i]=0;
    		else if (s[i]=='O') S[i]=1;
    		else S[i]=2;
    	}
    	dfs(1,0);
    	F[0][0][0]=1;
    	for (int i=0; i<n; i++){
    		for (int pre=0; pre<(1<<K); pre++)
    			for (int cas=0; cas<3; cas++)
    				F[(i+1)%2][pre][cas]=0;
    		for (int pre=0; pre<(1<<K); pre++)
    			for (int cas=0; cas<3; cas++)
    				if (F[i%2][pre][cas]){
    					for (int to=0; to<3; to++){
    						int now=To[pre][to],Tocas=cas;
    						if (to==cas) Tocas++;
    						else{
    							if (to==0) Tocas=1;
    							else Tocas=0;
    						}
    						if (Tocas==3) continue;
    						(F[(i+1)%2][now][Tocas]+=F[i%2][pre][cas])%=mod;
    					}
    				}
    	}
    	for (int now=0; now<(1<<K); now++) 
    		for (int cas=0; cas<3; cas++)
    			(ANS[ED[now]]+=F[n%2][now][cas])%=mod;
    	for (int i=0; i<=K; i++)
    		printf("%d
    ",ANS[i]);
    	return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/silenty/p/9816651.html
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