思路:
这题正解是dp
然后我打dp打挂了,改记搜就过了
转移时分三种情况就好啦
C o d e Code Code:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int a[2001],b[2001],f[2001][2001],k;
string s1,s2;
int dfs(int x,int y)
{
if(f[x][y] == -1)
{
if(x == 0) f[x][y] = y * k;else
if(y == 0) f[x][y] = x * k;else
f[x][y] =
min(dfs(x - 1, y - 1) + abs(a[x] - b[y]),
min (dfs (x - 1,y) + k, dfs(x,y - 1)+k));
//1,不填空格 此时f[x][y]=f[x-1][y-1]两边都用掉末尾的一个
//2,在A串末尾填空格,此时f[x][y]=f[x][y-1]空格抵消掉了B串末尾的一个
//3,在B串末尾填空格,可以自己理解一下
}
return f[x][y];
}
int main()
{
freopen ("blast.in","r",stdin);
freopen ("blast.out","w",stdout);
cin>>s1>>s2;
scanf("%d", &k);
int len1 = s1.length(), len2 = s2.length();
for(int i = 0; i < len1; ++i)
a[i + 1] = (int)s1[i];
for(int i = 0; i < len2; ++i)
b[i + 1] = (int)s2[i];
memset(f,-1,sizeof(f));
printf("%d",dfs(len1,len2));
return 0;
}