USACO 5.4 Character Recognition

Character Recognition

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 beginning of `font.in` (space and `a`).	Sample `charrec.in`, showing an `a` corrupted
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)

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……】

 /*

 ID: ivorysi

 LANG: C++

 PROG: charrec

 */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <queue>

 #include <set>

 #include <vector>

 #include <string.h>

 #include <cmath>

 #include <stack>

 #include <map>

 #define siji(i,x,y) for(int i=(x);i<=(y);++i)

 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j)

 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i)

 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j)

 #define inf 0x3f3f3f3f

 //#define ivorysi

 #define mo 97797977

 #define hash 974711

 #define base 47

 #define pss pair<string,string>

 #define MAXN 5000

 #define fi first

 #define se second

 #define pii pair<int,int>

 #define esp 1e-8

 typedef long long ll;

 using namespace std;

 char *os=" abcdefghijklmnopqrstuvwxyz?";

 char ch[][],word[][];

 int n1,n2;

 int c[][],b[],pr[][];

 void init() {

     FILE *fin=fopen("font.in","r");

     fscanf(fin,"%d",&n1);

     siji(i,,n1) {

         fscanf(fin,"%s",ch[i]+);

     }

     scanf("%d",&n2);

     siji(i,,n2) {

         scanf("%s",word[i]+);

     }

     siji(i,,n2) {

         b[i]=-;

         siji(j,,) {

             c[i][j]=-;

         }

     }

     b[]=;

 }

 int check(int d,int l) {

     if(c[d][l]!=-) return c[d][l];

     c[d][l]=inf;

     int t=,t1;

     int id1,id2;

     int ti=max(,l);

     siji(i,,) {

         t=;

         if((i+)*>n1) break;

         siji(j,,l) {

             siji(z,,) {

                 if(ch[i*+j][z]!=word[d+j][z])

                     ++t;

             }

         }

         if(l==) id1=-,id2=;

         if(l==) id1=,id2=-;

         siji(k,,) {

             if(l==) break;

             t1=;

             siji(j,,k-) {

                 siji(z,,) {

                     if(ch[i*+j][z]!=word[d+j][z]) ++t1;

                 }

             }

             siji(j,k+,ti) {

                 siji(z,,) {

                     if(ch[i*+j+id1][z]!=word[d+j+id2][z])

                         ++t1;

                 }

             }

             t=min(t,t1);

         }

         if(c[d][l]>t) {

             c[d][l]=t;

             pr[d][l]=i;

         }

     }

     if(c[d][l]>) {pr[d][l]=;}//假如损坏率超过30%

     return c[d][l];

 }

 int dfs(int m) {

     if(b[m]!=-) return b[m];

     b[m]=inf;

     siji(i,,) {

         if(m-i<) break;

         if(dfs(m-i)>=inf) continue;

         b[m]=min(b[m],dfs(m-i)+check(m-i,i));

     }

     return b[m];

 }

 void pra(int m) {

     if(m==) return;

     char cw;

     siji(i,,) {

         if(b[m]==b[m-i]+c[m-i][i]) {

             pra(m-i);

             cw=pr[m-i][i];

             break;

         }

     }

     printf("%c",os[cw]);

 }

 void solve() {

     init();

     dfs(n2);

     pra(n2);

     puts("");

 }

 int main(int argc, char const *argv[])

 {

 #ifdef ivorysi

     freopen("charrec.in","r",stdin);

     freopen("charrec.out","w",stdout);

 #else

     freopen("f1.in","r",stdin);

 #endif

     solve();

     return ;

 }