Luogu P2516 [HAOI2010]最长公共子序列 DP

首先$LIS$显然:$f[i][j]=max(f[i][j-1],f[i-1][j],(a[i]==b[j])*f[i-1][j-1])$

考虑如何转移数量：

首先，不管$a[i]$是否等于$b[j]$,

都有$h[i][j]+=h[i-1][j]*(f[i][j]==f[i-1][j])+h[i][j-1]*(f[i][j]==f[i][j-1])$

然后讨论$LIS$中第三种转移：

如果$a[i]==b[j]\ \&\&\ f[i][j]==f[i-1][j-1]+1$,有$h[i][j]+=h[i-1][j-1]$

而如果$a[i]!=b[j]\ \&\&\ f[i][j]==f[i-1][j-1]$,有$h[i][j]-=h[i-1][j-1]$,此处是考虑$h[i-1][j]$与$h[i][j-1]$对答案的重复转移，所以要减去；

#include<cstdio>
#include<iostream>
#include<cstring>
#define R register int
using namespace std;
namespace Fread {
    static char B[<<],*S=B,*D=B;
    #define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
    inline int g() {
        R ret=,fix=; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-:fix;
        do ret=ret*+(ch^); while(isdigit(ch=getchar())); return ret*fix;
    } inline bool isempty(const char& ch) {return ch<=||ch>=;}
    inline void gs(char* s) {register char ch; while(isempty(ch=getchar())); do *s++=ch; while(!isempty(ch=getchar()));}
}using Fread::g; using Fread::gs;
const int N=,M=1E+;
int n,m; char s1[N],s2[N];
int f[][N],h[][N];
signed main() {
    gs(s1+),gs(s2+); R l1=strlen(s1+)-,l2=strlen(s2+)-;
    R c=; for(R i=;i<=l2;++i) h[c][i]=;
    for(R i=;i<=l1;++i) {
        c^=; h[c][]=;
        for(R j=;j<=l2;++j) {
            h[c][j]=;
            f[c][j]=max(f[c^][j],f[c][j-]);
            if(s1[i]==s2[j]) f[c][j]=max(f[c][j],f[c^][j-]+);
            if(f[c][j]==f[c^][j]) h[c][j]+=h[c^][j];
            if(f[c][j]==f[c][j-]) h[c][j]+=h[c][j-];
            if(s1[i]==s2[j]&&f[c][j]==f[c^][j-]+) h[c][j]+=h[c^][j-];
            if(s1[i]!=s2[j]&&f[c][j]==f[c^][j-]) h[c][j]-=h[c^][j-];
            h[c][j]=(h[c][j]+M)%M;
        }
    } printf("%d\n%d\n",f[c][l2],h[c][l2]);
}

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2019.07.11