ZOJ 3697 Bad-written Number

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dp[ i ][ j ] += dp[ i-1 ][ k ]，match(j,k). dp[ i ][ j ]表示第 i 为 j 时的方案数。

与其说是DP，不如说是模拟题。

第一个和最后一个数字要单独讨论，中间的要符合剩下的条件：

中间一列和剩下的两列中的 ‘| ’要全部符合输入状态。且衔接部分要取 '|' 的并集，且并集要与输入状态相同。

第一个数字要前两列全部符合输入状态，最后一个数则要后两列。

为毛一到周赛就各种不会 -_- !

#include <iostream>#include <algorithm>#include <cstdlib>#include <cstdio>#include <cstring>#include <queue>#include <cmath>#include <stack>#pragma comment(linker, "/STACK:1024000000");#define EPS (1e-8)#define LL long long int#define ULL unsigned long long int#define _LL __int64#define _INF 0x3f3f3f3f#define INF 40000000#define Mod 1000000007using namespace std;char s[3][20010];int dp[10010][10];char dig[10][3][4] ={    {{' ','_',' ','\0'},{'|',' ','|','\0'},{'|','_','|','\0'}},    {{' ',' ',' ','\0'},{' ',' ','|','\0'},{' ',' ','|','\0'}},    {{' ','_',' ','\0'},{' ','_','|','\0'},{'|','_',' ','\0'}},    {{' ','_',' ','\0'},{' ','_','|','\0'},{' ','_','|','\0'}},    {{' ',' ',' ','\0'},{'|','_','|','\0'},{' ',' ','|','\0'}},    {{' ','_',' ','\0'},{'|','_',' ','\0'},{' ','_','|','\0'}},    {{' ','_',' ','\0'},{'|','_',' ','\0'},{'|','_','|','\0'}},    {{' ','_',' ','\0'},{' ',' ','|','\0'},{' ',' ','|','\0'}},    {{' ','_',' ','\0'},{'|','_','|','\0'},{'|','_','|','\0'}},    {{' ','_',' ','\0'},{'|','_','|','\0'},{' ','_','|','\0'}},};bool Match(int w){    int i ,j;    for(i = 0;i < 3; ++i)    {        for(j = 0;j  < 3; ++j)        {            if(dig[w][i][j] != s[i][j+1])                return false;        }    }    return true;}bool Match_H(int w){    int i,j;    for(i = 0;i < 3; ++i)    {        for(j = 0;j < 3; ++j)        {            if(dig[w][i][j] != s[i][j+1] && (j != 2 || (j == 2 && dig[w][i][j] == '|')))                return false;        }    }    return true;}bool Match_(int len,int v){    int i,j;    for(i = 0;i < 3; ++i)    {        for(j = 0;j < 3; ++j)        {            if(dig[v][i][j] != s[i][j+len] && (j != 2 || (j == 2 && dig[v][i][j] == '|')) && (j != 0 || (j == 0 && dig[v][i][j] == '|')) )                return false;        }    }    return true;}bool Match_M(int len,int u,int v){    int i;    for(i = 0;i < 3; ++i)    {        if(dig[u][i][2] != '|' && dig[v][i][0] != '|' && s[i][len] == '|')            return false;        if(dig[u][i][2] != ' ' && dig[v][i][0] != ' ' && s[i][len] == ' ')            return false;    }    return true;}bool Match_E(int len,int u,int v){    int i,j;    for(i = 0;i < 3; ++i)    {        if(dig[u][i][2] != '|' && dig[v][i][0] != '|' && s[i][len] == '|')            return false;        if(dig[u][i][2] != ' ' && dig[v][i][0] != ' ' && s[i][len] == ' ')            return false;    }    for(i = 0;i < 3; ++i)    {        for(j = 0;j < 3; ++j)        {            if(dig[v][i][j] != s[i][j+len] && (j != 0 || (j == 0 && dig[v][i][j] == '|')) )                return false;        }    }    return true;}void solve(int n){    int i,j,k,l;    if(n == 1)    {        for(i = 0;i <= 9; ++i)        {            if(Match(i))                break;        }        if(i == 10)        {            printf("0\n");        }        else        {            printf("1\n");        }        return ;    }    for(i = 0;i <= n; ++i)    {        memset(dp[i],0,sizeof(dp[i]));    }    for(i = 0;i <= 9; ++i)    {        if(Match_H(i))            dp[1][i] = 1;        else            dp[1][i] = 0;    }    int Top = 2*n+1 - 2;    int a,b;    for(i = 3;i < Top; i += 2)    {        a = i/2 + 1;        b = a - 1;        for(j = 0;j <= 9; ++j)        {            if(Match_(i,j))            {                for(k = 0;k <= 9; ++k)                {                    if(dp[b][k] && Match_M(i,k,j))                    {                        dp[a][j] += dp[b][k];                        dp[a][j] %= Mod;                    }                }            }        }        for(l = 0;l <= 9; ++l)        {            if(dp[a][l])                break;        }        if(l == 10)        {            printf("0\n");            return ;        }    }    a = i/2+1;    b = a-1;    for(i = 0;i <= 9; ++i)    {        for(j = 0;j <= 9; ++j)        {            if(dp[b][j] && Match_E(Top,j,i))            {                dp[a][i] += dp[b][j];                dp[a][i] %= Mod;            }        }    }    int ans = 0;    for(i = 0;i <= 9; ++i)    {        ans += dp[n][i];        ans %= Mod;    }    printf("%d\n",ans);    return ;}int main(){    int i,T,n;    scanf("%d",&T);    while(T--)    {        scanf("%d%*c",&n);        for(i = 0; i < 3; ++i)        {            gets(s[i]+1);        }        solve(n);    }    return 0;}

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