This problem requires you to write a program that performs character recognition.
Each ideal character image has 20 lines of 20 digits. Each digit is a `0' or a `1'. See Figure 1a (way below) for the layout of character images in the file.
The file font.in contains representations of 27 ideal character images in this order:
_abcdefghijklmnopqrstuvwxyz
where _ represents the space character. Each ideal character is 20 lines long.
The input file contains one or more potentially corrupted character images. A character image might be corrupted in these ways:
- at most one line might be duplicated (and the duplicate immediately follows)
- at most one line might be missing
- some 0's might be changed to 1's
- some 1's might be changed to 0's.
No character image will have both a duplicated line and a missing line. No more than 30% of the 0's and 1's will be changed in any character image in the evaluation datasets.
In the case of a duplicated line, one or both of the resulting lines may have corruptions, and the corruptions may be different.
Write a program to recognize the sequence of one or more characters in the image provided in the input file using the font provided in file font.in.
Recognize a character image by choosing the font character images that require the smallest number of overall changed 1's and 0's to be corrupted to the given font image, given the most favourable assumptions about duplicated or omitted lines. Count corruptions in only the least corrupted line in the case of a duplicated line. You must determine the sequence of characters that most closely matches the input sequence (the one that requires the least number of corruptions). There is a unique best solution for each evaluation dataset.
A correct solution will use precisely all of the data supplied in the input file.
PROGRAM NAME: charrec
INPUT FORMAT (both input files)
Both input files begin with an integer N (19 <= N < 1200) that specifies the number of lines that follow:
N
(digit1)(digit2)(digit3) ... (digit20)
(digit1)(digit2)(digit3) ... (digit20)
...
Each line of data is 20 digits wide. There are no spaces separating the zeros and ones.
The file font.in describes the font. It will always contain 541 lines. It may differ for each evaluation dataset.
SAMPLE INPUT (file charrec.in)
|
Incomplete sample showing the |
Sample |
|
font.in |
charrec.in |
| 540 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000011100000000000 00000111111011000000 00001111111001100000 00001110001100100000 00001100001100010000 00001100000100010000 00000100000100010000 00000010000000110000 00000001000001110000 00001111111111110000 00001111111111110000 00001111111111000000 00001000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 00000000000000000000 |
19 00000000000000000000 00000000000000000000 00000000000000000000 00000011100000000000 00100111011011000000 00001111111001100000 00001110001100100000 00001100001100010000 00001100000100010000 00000100000100010000 00000010000000110000 00001111011111110000 00001111111111110000 00001111111111000000 00001000010000000000 00000000000000000000 00000000000001000000 00000000000000000000 00000000000000000000 |
| Figure 1a | Figure 1b |
OUTPUT FORMAT
Your program must produce an output file that contains a single string of the characters recognized. Its format is a single line of ASCII text. The output should not contain any separator characters. If your program does not recognize a particular character, it must output a ? in the appropriate position.
SAMPLE OUTPUT (file charrec.out)
a
Note that the 'sample output' formerly displayed a blank followed by an 'a', but that seems wrong now.
————————————————————————————————————————题解
任何不会做的题一定不能虚的坦然自若地说完题解
这道题写模拟一定会超时，然后我们发现有最优子结构
也就是dp[m]表示匹配到了第m行最小的不相符字符数，我们再开一个数组c[j][k]表示从第j行下匹配了k行的最小代价，同时记录这个字符是什么
20行直接匹配，19行枚举pass掉该字符标准表示的哪一行，21行枚举pass掉匹配的这21行的哪一行然后匹配
然后记搜转移，因为并不是所有状态都要用到，最后一个点是0.9s才过
【把font.in点开后粘到编辑器然后看缩略图会发现极为良心的……当然近视眼只需要摘下眼镜了……】
1 /* 2 ID: ivorysi 3 LANG: C++ 4 PROG: charrec 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstring> 9 #include <algorithm> 10 #include <queue> 11 #include <set> 12 #include <vector> 13 #include <string.h> 14 #include <cmath> 15 #include <stack> 16 #include <map> 17 #define siji(i,x,y) for(int i=(x);i<=(y);++i) 18 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j) 19 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i) 20 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j) 21 #define inf 0x3f3f3f3f 22 //#define ivorysi 23 #define mo 97797977 24 #define hash 974711 25 #define base 47 26 #define pss pair<string,string> 27 #define MAXN 5000 28 #define fi first 29 #define se second 30 #define pii pair<int,int> 31 #define esp 1e-8 32 typedef long long ll; 33 using namespace std; 34 char *os=" abcdefghijklmnopqrstuvwxyz?"; 35 char ch[545][25],word[1205][25]; 36 int n1,n2; 37 int c[1205][25],b[1205],pr[1205][25]; 38 void init() { 39 FILE *fin=fopen("font.in","r"); 40 fscanf(fin,"%d",&n1); 41 siji(i,1,n1) { 42 fscanf(fin,"%s",ch[i]+1); 43 } 44 scanf("%d",&n2); 45 siji(i,1,n2) { 46 scanf("%s",word[i]+1); 47 } 48 siji(i,0,n2) { 49 b[i]=-1; 50 siji(j,19,21) { 51 c[i][j]=-1; 52 } 53 } 54 b[0]=0; 55 } 56 int check(int d,int l) { 57 if(c[d][l]!=-1) return c[d][l]; 58 c[d][l]=inf; 59 int t=0,t1; 60 int id1,id2; 61 int ti=max(20,l); 62 siji(i,0,26) { 63 t=0; 64 if((i+1)*20>n1) break; 65 siji(j,1,l) { 66 siji(z,1,20) { 67 if(ch[i*20+j][z]!=word[d+j][z]) 68 ++t; 69 } 70 } 71 if(l==21) id1=-1,id2=0; 72 if(l==19) id1=0,id2=-1; 73 siji(k,1,20) { 74 if(l==20) break; 75 t1=0; 76 siji(j,1,k-1) { 77 siji(z,1,20) { 78 if(ch[i*20+j][z]!=word[d+j][z]) ++t1; 79 } 80 } 81 siji(j,k+1,ti) { 82 siji(z,1,20) { 83 if(ch[i*20+j+id1][z]!=word[d+j+id2][z]) 84 ++t1; 85 } 86 } 87 t=min(t,t1); 88 } 89 if(c[d][l]>t) { 90 c[d][l]=t; 91 pr[d][l]=i; 92 } 93 } 94 if(c[d][l]>120) {pr[d][l]=27;}//假如损坏率超过30% 95 return c[d][l]; 96 } 97 98 int dfs(int m) { 99 if(b[m]!=-1) return b[m]; 100 b[m]=inf; 101 siji(i,19,21) { 102 if(m-i<0) break; 103 if(dfs(m-i)>=inf) continue; 104 b[m]=min(b[m],dfs(m-i)+check(m-i,i)); 105 } 106 return b[m]; 107 } 108 void pra(int m) { 109 if(m==0) return; 110 char cw; 111 siji(i,19,21) { 112 if(b[m]==b[m-i]+c[m-i][i]) { 113 pra(m-i); 114 cw=pr[m-i][i]; 115 break; 116 } 117 } 118 printf("%c",os[cw]); 119 } 120 void solve() { 121 init(); 122 dfs(n2); 123 pra(n2); 124 puts(""); 125 } 126 int main(int argc, char const *argv[]) 127 { 128 #ifdef ivorysi 129 freopen("charrec.in","r",stdin); 130 freopen("charrec.out","w",stdout); 131 #else 132 freopen("f1.in","r",stdin); 133 #endif 134 solve(); 135 return 0; 136 }