【jzoj5237】【GDOI2018模拟8.7】【最长公共子序列 】【动态规划】

题目大意

解题思路

设f[i][j]表示a考虑前i个字符，b考虑前j个字符的lcs，g[i][j]表示长度为f[i][j]的匹配的个数。g[i][j]可以由g[i-1][j],g[i][j-1],g[i-1][j-1]中f相同的转移过来，但是如果f[i][j]==f[i-1][j]==f[i][j-1]==f[i-1][j-1]，g[i][j]=g[i-1][j]+g[i][j-1]-g[i-1][j-1]即可。

code

#include
#include
#include
#include
#define LF double
#define LL long long
#define ULL unsigned int
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=5000+9,mo=1e9+7;
int n,m,f[mn][mn],g[mn][mn];
char a[mn],b[mn];
int max(int x,int y){
    return (x>y)?x:y;
}
int main(){
    //freopen("lcs.in","r",stdin);
    //freopen("lcs.out","w",stdout);
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    scanf("%s\n",a+1);n=strlen(a+1);
    scanf("%s\n",b+1);m=strlen(b+1);
    g[0][0]=1;
    fo(i,1,n)g[i][0]=1;
    fo(i,1,m)g[0][i]=1;
    fo(i,1,n)fo(j,1,m){
        f[i][j]=max(max(f[i-1][j],f[i][j-1]),f[i-1][j-1]+(a[i]==b[j]));
        if(f[i][j]==f[i-1][j])g[i][j]=(g[i][j]+g[i-1][j])%mo;
        if(f[i][j]==f[i][j-1])g[i][j]=(g[i][j]+g[i][j-1])%mo;
        if(f[i][j]==f[i-1][j-1]+(a[i]==b[j]))g[i][j]=(g[i][j]+g[i-1][j-1])%mo;
        if((f[i][j]==f[i-1][j])&&(f[i][j]==f[i][j-1])&&(f[i][j]==f[i-1][j-1]))
            g[i][j]=(g[i][j]-g[i-1][j-1]*2)%mo;
    }
    printf("%d\n%d",f[n][m],(g[n][m]+mo)%mo);
    return 0;
}