思路：设字符串x的长度为n，y的长度为m，那么答案一定在[0, m]之间，那么可以二分求答案。

d(i, j)表示第一个串前i个字符至少需要经过多少次才能的到第二个串的前j个字符，转移方程d(i, j) = min{d(i-1, j-1) + is_same(a[i], b[j]), d(i-1, j) + 1, d(i, j-1) + 1};

如何判断某个答案mid是否合理？当a串中连续k个字符能够在mid次操作类转换成第二个串，就把d(k, 0)变成0即可。

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 5000 + 5;
char x[maxn], y[55];
int d[maxn][55], n, m;
bool is_ok(int len) {
	d[0][0] = 0;
	for(int i = 1; i <= n; ++i) d[i][0] = i;
	for(int i = 1; i <= m; ++i) d[0][i] = i;

	for(int i = 1; i <= n; ++i){
		for(int j = 1; j <= m; ++j) {
			d[i][j] = d[i-1][j-1] + (x[i] == y[j] ? 0 : 1);
			d[i][j] = min(d[i][j], d[i-1][j] + 1);
			d[i][j] = min(d[i][j], d[i][j-1] + 1);
		}
		if(d[i][m] <= len) d[i][0] = 0;
	}

	return d[n][m] <= len;
}

int bin_search(int l, int r) {
	while(l < r) {
		int mid = (l + r) / 2;
		if(is_ok(mid)) r = mid;
		else l = mid + 1;
	}
	return r;
}

int main() {
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%s%s", y+1, x+1);
		n = strlen(x+1), m = strlen(y+1);
		printf("%d\n", bin_search(0, m));
	}
	return 0;
} 

如有不当之处欢迎指出！