一、內容
Biologists finally invent techniques of repairing DNA that contains segments causing kinds of inherited diseases. For the sake of simplicity, a DNA is represented as a string containing characters 'A', 'G' , 'C' and 'T'. The repairing techniques are simply to change some characters to eliminate all segments causing diseases. For example, we can repair a DNA "AAGCAG" to "AGGCAC" to eliminate the initial causing disease segments "AAG", "AGC" and "CAG" by changing two characters. Note that the repaired DNA can still contain only characters 'A', 'G', 'C' and 'T'.
You are to help the biologists to repair a DNA by changing least number of characters.
Input
The input consists of multiple test cases. Each test case starts with a line containing one integers N (1 ≤ N ≤ 50), which is the number of DNA segments causing inherited diseases.
The following N lines gives N non-empty strings of length not greater than 20 containing only characters in "AGCT", which are the DNA segments causing inherited disease.
The last line of the test case is a non-empty string of length not greater than 1000 containing only characters in "AGCT", which is the DNA to be repaired.
The last test case is followed by a line containing one zeros.
Output
For each test case, print a line containing the test case number( beginning with 1) followed by the
number of characters which need to be changed. If it's impossible to repair the given DNA, print -1.
Sample Input
2
AAA
AAG
AAAG
2
A
TG
TGAATG
4
A
G
C
T
AGT
0
Sample Output
Case 1: 1
Case 2: 4
Case 3: -1
二、思路
三、代碼
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int N = 1005;
int n, tr[N][4], ne[N], f[N][N], cnt[N], len;
char s[N];
int getC(char c) {
if (c == 'A') return 0;
if (c == 'T') return 1;
if (c == 'G') return 2;
return 3;
}
void add() {
int p = 0;
for (int i = 0; s[i]; i++) {
int j = getC(s[i]);
if (!tr[p][j]) tr[p][j] = len++;
p = tr[p][j];
}
cnt[p] = 1; //標記這個字符串
}
void build() {
queue<int> q;
for (int i = 0; i < 4; i++) { //將第二層的節點入隊
if (tr[0][i]) q.push(tr[0][i]);
}
while (!q.empty()) {
int p = q.front(); q.pop();
for (int j = 0; j < 4; j++) {
int c = tr[p][j]; //子節點
if (!c) tr[p][j] = tr[ne[p]][j];
else {
ne[c] = tr[ne[p]][j];
cnt[c] |= cnt[ne[c]]; //把前面存在的字符串狀態加上
q.push(c);
}
}
}
}
int main() {
int T = 1;
while (scanf("%d", &n), n) {
memset(cnt, 0, sizeof(cnt));
memset(ne, 0, sizeof(ne)); len = 1;
memset(f, 0x3f, sizeof(f));
memset(tr, 0, sizeof(tr));
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%s", s); add();
}
build();
scanf("%s", s + 1); n = strlen(s + 1);
f[0][0] = 0;
for (int i = 1; i <= n; i++) {
int c = getC(s[i]);
for (int j = 0; j <= len; j++) {
for (int k = 0; k < 4; k++) {
int t = (c != k); //代表需要修改的次數 若i字符與k不相同則爲1
int tj = tr[j][k];
//代表沒有出現題目所給的串
if (!cnt[tj]) f[i][tj] = min(f[i][tj], f[i - 1][j] + t);
}
}
}
int ans = 1e9;
for (int j = 0; j <= len; j++) ans = min(ans, f[n][j]);
printf("Case %d: %d\n", T++, ans == 1e9 ? -1 : ans);
}
return 0;
}