51nod 编辑距离 + 滚动数组优化

这道题一开始觉得增加和删除会移动字符串的位置很不好做

两个字符串dp状态一般是第一个前i个和第二个前j个

#include<cstdio>
#include<algorithm>
#include<cstring>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;

const int MAXN = 1123;
char a[MAXN], b[MAXN];
int f[MAXN][MAXN];

int main()
{
	scanf("%s%s", a + 1, b + 1);
	int n = strlen(a + 1), m = strlen(b + 1);
	REP(i, 0, n + 1) f[i][0] = i;
	REP(i, 0, m + 1) f[0][i] = i;
	REP(i, 1, n + 1)
		REP(j, 1, m + 1)
		{
			if(a[i] == b[j]) f[i][j] = f[i-1][j-1];
			else f[i][j] = min(f[i-1][j], min(f[i][j-1], f[i-1][j-1])) + 1;
		}
	printf("%d
", f[n][m]);
	return 0;
}

这里的转移只和上一行有关
所以可以用滚动数组优化空间
处理过的就是这一行，没处理过的就是上一行

#include<cstdio>
#include<algorithm>
#include<cstring>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;
 
const int MAXN = 1123;
char a[MAXN], b[MAXN];
int f[MAXN];
 
int main()
{
	scanf("%s%s", a + 1, b + 1);
	int n = strlen(a + 1), m = strlen(b + 1);
	REP(i, 1, m + 1) f[i] = i;
	REP(i, 1, n + 1)
	{
		int last = i - 1; //f[i-1][0] = i - 1
		f[0] = i;  //f[i][0] = i
		REP(j, 1, m + 1)
		{
			int temp = f[j]; 
			if(a[i] == b[j]) f[j] = last; //last = f[i-1][j-1]
			else f[j] = min(f[j], min(f[j-1], last)) + 1;
			last = temp;
		}
	}
	printf("%d
", f[m]);
	return 0;
}