[CodeM初赛A轮]B

来源：互联网发布：功夫软件编辑：程序博客网时间：2024/05/21 00:52

题解

设dp f[i1,i2,j1,j2]表示选出A串的[i1,i2]以及B串的[j1,j2]能否组成回文串。
dp转移显然。
但是边界比较麻烦。
考虑先做dp fa[i,j]和fb[i,j]分表表示A串的[i,j]和B串的[i,j]是不是回文串，然后瞎枚举弄出一些f的初值。
回文串完全在一个串中也要判断。
有点坑QAQ
详见代码

#include<cstdio>#include<cstring>#include<algorithm>#define fo(i,a,b) for(i=a;i<=b;i++)#define fd(i,a,b) for(i=a;i>=b;i--)using namespace std;const int maxn=50+10,inf=100000000;char a[maxn],b[maxn];int fa[maxn][maxn],fb[maxn][maxn],f[maxn][maxn][maxn][maxn];int i,j,i1,i2,jj1,j2,k,l,t,n,m,ca,ans;int main(){    scanf("%d",&ca);    while (ca--){        scanf("%s",a+1);        scanf("%s",b+1);        n=strlen(a+1);m=strlen(b+1);        fo(i,0,n)            fo(j,0,n)                fa[i][j]=-inf;        fo(i,0,m)            fo(j,0,m)                fb[i][j]=-inf;        fo(i1,0,n)            fo(i2,0,n)                fo(jj1,0,m)                    fo(j2,0,m)                        f[i1][i2][jj1][j2]=-inf;        fo(i,1,n)            fo(j,1,m)                if (a[i]==b[j]) f[i][i][j][j]=2;        ans=0;        fo(i,1,n) fa[i][i]=1;        fo(i,1,n-1)             if (a[i]==a[i+1]) fa[i][i+1]=2;        fd(i,n,1)            fo(j,1,n)                if (fa[i][j]!=-inf){                    t=fa[i][j];                    ans=max(ans,t);                    if (i>1&&j<n&&a[i-1]==a[j+1]) fa[i-1][j+1]=t+2;                    fo(k,1,m){                        if (j<n&&b[k]==a[j+1]) f[i][j+1][k][k]=t+2;                        if (i>1&&b[k]==a[i-1]) f[i-1][j][k][k]=t+2;                        if (k<m&&b[k]==b[k+1]) f[i][j][k][k+1]=t+2;                    }                }        fo(i,1,m) fb[i][i]=1;        fo(i,1,m-1)             if (b[i]==b[i+1]) fb[i][i+1]=2;        fd(i,m,1)            fo(j,1,m)                if (fb[i][j]!=-inf){                    t=fb[i][j];                    ans=max(ans,t);                    if (i>1&&j<m&&b[i-1]==b[j+1]) fb[i-1][j+1]=t+2;                    fo(k,1,n){                        if (j<m&&a[k]==b[j+1]) f[k][k][i][j+1]=t+2;                        if (i>1&&a[k]==b[i-1]) f[k][k][i-1][j]=t+2;                        if (k<n&&a[k]==a[k+1]) f[k][k+1][i][j]=t+2;                    }                }        fd(i1,n,1)            fo(i2,1,n)                fd(jj1,m,1)                    fo(j2,1,m)                        if (f[i1][i2][jj1][j2]!=-inf){                            t=f[i1][i2][jj1][j2];                            ans=max(ans,t);                            if (ans==6){                                t=t;                            }                            if (i1>1&&i2<n&&a[i1-1]==a[i2+1]) f[i1-1][i2+1][jj1][j2]=t+2;                            if (jj1>1&&j2<m&&b[jj1-1]==b[j2+1]) f[i1][i2][jj1-1][j2+1]=t+2;                            if (i1>1&&j2<m&&a[i1-1]==b[j2+1]) f[i1-1][i2][jj1][j2+1]=t+2;                            if (i2<n&&jj1>1&&b[jj1-1]==a[i2+1]) f[i1][i2+1][jj1-1][j2]=t+2;                        }        printf("%d\n",ans);    }}

阅读全文

0 0