pku 3356 AGTC

题目:

http://acm.pku.edu.cn/JudgeOnline/problem?id=3356

 

#include "stdio.h"//f(i,j)表示字符串A的前i个字符到字符串B的前j个字符匹配所需的最小改变步数int f[1002][1002];int min(int a,int b,int c){ int temp; temp = a <= b ? a : b; return c <= temp ? c : temp;}int main(){ int i,j; int len1,len2; char str1[1002],str2[1002]; while(scanf("%d%s",&len1,str1 + 1)!=EOF){scanf("%d%s",&len2,str2 + 1);//注意初始化for(i=0;i<=len1;i++) f[i][0]=i; for(j=0;j<=len2;j++) f[0][j]=j; for(i=1;i<=len1;i++) for(j=1;j<=len2;j++) {//不改变...改变A...改变Bif(str1[i]==str2[j])f[i][j]=min(f[i-1][j-1],f[i-1][j]+1,f[i][j-1]+1); else f[i][j]=min(f[i-1][j-1]+1,f[i-1][j]+1,f[i][j-1]+1); } printf("%d/n",f[len1][len2]);}return 0;}