usaco5.4.2 Character Recognition
一 原题
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)
aNote that the 'sample output' formerly displayed a blank followed by an 'a', but that seems wrong now.
二 分析
三 代码
USER: Qi Shen [maxkibb3] TASK: charrec LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.022 secs, 6976 KB] Test 2: TEST OK [0.011 secs, 6976 KB] Test 3: TEST OK [0.022 secs, 6976 KB] Test 4: TEST OK [0.032 secs, 6976 KB] Test 5: TEST OK [0.076 secs, 6976 KB] Test 6: TEST OK [0.151 secs, 6976 KB] Test 7: TEST OK [0.194 secs, 6976 KB] Test 8: TEST OK [0.194 secs, 6976 KB] Test 9: TEST OK [0.162 secs, 6976 KB] All tests OK.
YOUR PROGRAM ('charrec') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.
/*
ID:maxkibb3
LANG:C++
PROB:charrec
*/
#include<cstdio>
const int INF = 1e6;
const int LETTERNUM = 27;
const int LINENUM = 20;
const int MAXLINE = 1205;
const int MAGICNUM = 22;
const char A[] = " abcdefghijklmnopqrstuvwxyz";
int N;
char F[LETTERNUM][LINENUM][LINENUM + 1];
char S[MAXLINE][LINENUM + 1];
// For DP
int D[LETTERNUM][LINENUM][MAXLINE];
int C[MAXLINE][MAGICNUM]; int From[MAXLINE][MAGICNUM];
int Dp[MAXLINE]; int Ffrom[MAXLINE];
void init() {
freopen("font.in", "r", stdin);
int tmp;
scanf("%d", &tmp);
for(int i = 0; i < tmp; i++) {
getchar();
scanf("%s", F[i / LINENUM][i % LINENUM]);
}
freopen("charrec.in", "r", stdin);
scanf("%d", &N);
for(int i = 0; i < N; i++) {
getchar();
scanf("%s", S[i]);
}
freopen("charrec.out", "w", stdout);
}
void set_diff() {
for(int i = 0; i < LETTERNUM; i++)
for(int j = 0; j < LINENUM; j++)
for(int k = 0; k < N; k++) {
int tot = 0;
for(int l = 0; l < LINENUM; l++)
tot += (F[i][j][l] != S[k][l]);
D[i][j][k] = tot;
}
}
void set_cost() {
for(int i = 0; i < N; i++)
C[i][19] = C[i][20] = C[i][21] = INF;
int i, j, k, l;
// for 19-case
for(i = 18; i < N; i++)
for(j = 0; j < LETTERNUM; j++)
for(k = 0; k < LINENUM; k++) {// choose the deleted line
int tot = 0;
for(l = 0; l < LINENUM - 1; l++) {
if(l >= k) tot += D[j][l + 1][i - 18 + l];
else tot += D[j][l][i - 18 + l];
}
if(tot < C[i][19]) C[i][19] = tot, From[i][19] = j;
}
// for 20-case
for(i = 19; i < N; i++)
for(j = 0; j < LETTERNUM; j++) {
int tot = 0;
for(k = 0; k < LINENUM; k++)
tot += D[j][k][i - 19 + k];
if(tot < C[i][20]) C[i][20] = tot, From[i][20] = j;
}
// for 21-case
for(i = 20; i < N; i++)
for(j = 0; j < LETTERNUM; j++)
for(k = 0; k < LINENUM; k++) { // choose the duplicated line
int tot = 0;
for(int l = 0; l < LINENUM; l++) {
if(l >= k) tot += D[j][l][i - 20 + l + 1];
else tot += D[j][l][i - 20 + l];
}
if(tot < C[i][21]) C[i][21] = tot, From[i][21] = j;
}
}
void dp() {
for(int i = 0; i < N; i++) Dp[i] = INF;
for(int i = 18; i < N; i++) {
int prev = (i < 19) ? 0 : Dp[i - 19];
if(prev + C[i][19] < Dp[i]) {
Dp[i] = prev + C[i][19];
Ffrom[i] = 19;
}
prev = (i < 20) ? 0 : Dp[i - 20];
if(prev + C[i][20] < Dp[i]) {
Dp[i] = prev + C[i][20];
Ffrom[i] = 20;
}
prev = (i < 21) ? 0 : Dp[i - 21];
if(prev + C[i][21] < Dp[i]) {
Dp[i] = prev + C[i][21];
Ffrom[i] = 21;
}
}
}
void print(int _idx) {
if(_idx == -1) return;
print(_idx - Ffrom[_idx]);
printf("%c", A[From[_idx][Ffrom[_idx]]]);
}
void solve() {
set_diff();
set_cost();
dp();
print(N - 1);
printf("\n");
}
int main() {
init();
solve();
return 0;
}