[BZOJ4565][HAOI2016]字符合并(区间状压DP)

https://blog.csdn.net/xyz32768/article/details/81591955

首先区间DP和状压DP是比较明显的，设f[L][R][S]为将[L,R]这一段独立操作最终得到的字符序列为S的最大收益。其中S的位数为(R-L)%(k-1)+1。

枚举R第一次参与的操作的左端点mid与这次操作得到的数转移，若当前长度为k则要加上当前操作收益。

注意这个mid和R的距离一定是k-1的倍数，这样总状态和转移就很少，于是可以$O(n^32^8)$通过。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=310;
const ll inf=1e14;
int n,k,a[N],c[N],w[N];
ll ans,f[N][N][N];
char s[N];

int main(){
    freopen("bzoj4565.in","r",stdin);
    freopen("bzoj4565.out","w",stdout);
    scanf("%d%d%s",&n,&k,s+1);
    rep(i,1,n) a[i]=s[i]-'0';
    for (int i=0; i<(1<<k); i++) scanf("%d%d",&c[i],&w[i]);
    rep(i,0,n) rep(j,0,n) rep(z,0,(1<<k)) f[i][j][z]=-inf;
    rep(i,1,n) f[i][i][a[i]]=0;
    rep(l,2,n) rep(i,1,n-l+1){
        int j=i+l-1,len=j-i;
        while (len>k-1) len-=k-1;
        for (int mid=j; mid>i; mid-=k-1)
            for (int z=0; z<(1<<len); z++){
                    f[i][j][z<<1]=max(f[i][j][z<<1],f[i][mid-1][z]+f[mid][j][0]);
                    f[i][j][(z<<1)|1]=max(f[i][j][(z<<1)|1],f[i][mid-1][z]+f[mid][j][1]);
                }
        if (len==k-1){
            ll g[2]; g[0]=g[1]=-inf;
            for (int z=0; z<(1<<k); z++) g[c[z]]=max(g[c[z]],f[i][j][z]+w[z]);
            f[i][j][0]=g[0]; f[i][j][1]=g[1];
        }
    }
    for (int i=0; i<(1<<k); i++) ans=max(ans,f[1][n][i]);
    printf("%lld
",ans);
    return 0;
}