输出三行,第一行和第二行均为一行字符串,分别表示两个字符串str1,str2。。第三行为三个正整数,代表ic,dc和rc。(1<=ic<=10000、1<=dc<=10000、1<=rc<=10000)
输出一个整数,表示编辑的最小代价。
abc adc 5 3 2
2
abc adc 5 3 100
8
abc abc 5 3 2
0
时间复杂度,空间复杂度
。(n,m代表两个字符串长度)
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
String str1 = sc.nextLine().trim();
String str2 = sc.nextLine().trim();
String[] s = sc.nextLine().trim().split(" ");
int ic=Integer.parseInt(s[0]);
int dc=Integer.parseInt(s[1]);
int rc=Integer.parseInt(s[2]);
int res=new Solution().getMinCost(str1,str2,ic,dc,rc);
System.out.println(res);
}
}
}
class Solution {
public int getMinCost(String str1,String str2,int ic,int dc,int rc){
int[][] dp=new int[str1.length()][str2.length()];
boolean flag=false;
for (int i = 0; i < str1.length(); i++) {
if(!flag&&str1.charAt(i)==str2.charAt(0)) flag=true;
if(flag){
dp[i][0]=i*dc;
}else{
dp[i][0]=Math.min(i*dc+rc,(i+1)*dc+ic);
}
}
flag=false;
for (int i = 0; i < str2.length(); i++) {
if(!flag&&str2.charAt(i)==str1.charAt(0)) flag=true;
if(flag){
dp[0][i]=i*ic;
}else{
dp[0][i]=Math.min(i*ic+rc,(i+1)*ic+dc);
}
}
for (int i = 1; i < str1.length(); i++) {
for (int j = 1; j < str2.length(); j++) {
if(str1.charAt(i)==str2.charAt(j)){
dp[i][j]=dp[i-1][j-1];
}else{
dp[i][j]=Math.min(dp[i-1][j]+dc,dp[i][j-1]+ic);
dp[i][j]=Math.min(dp[i][j],dp[i-1][j-1]+rc);
dp[i][j]=Math.min(dp[i][j],dp[i-1][j-1]+ic+dc);
}
}
}
return dp[str1.length()-1][str2.length()-1];
}
} import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str1 = sc.nextLine();
String str2 = sc.nextLine();
int ic = sc.nextInt();
int dc = sc.nextInt();
int rc = sc.nextInt();
System.out.println(minCount(str1, str2, ic, dc, rc));
}
public static int minCount(String str1, String str2, int ic, int dc, int rc) {
int len1 = str1.length();
int len2 = str2.length();
int[][] dp = new int[len1 + 1][len2 + 1];
dp[0][0] = 0;
for (int j = 1; j <= len2; j++) {
dp[0][j] = j * ic;
}
for (int i = 1; i <= len1; i++) {
dp[i][0] = i * dc;
}
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (str1.charAt(i - 1) == str2.charAt(j - 1))
dp[i][j] = dp[i - 1][j - 1];
else
dp[i][j] = min(dp[i - 1][j - 1] + rc, dp[i - 1][j] + dc, dp[i][j - 1] + ic);
}
}
return dp[len1][len2];
}
public static int min(int a, int b, int c) {
int temp = a;
if (b < temp) temp = b;
if (c < temp) temp = c;
return temp;
}
}