需要注意c[i][j]的递推算法和sp,sq,ep,eq的赋初值。由于dp[i][j]表示第一个字符串取i个,第二个字符串取j个,所以不论dp[i][j]从dp[i-1][j]+c[i-1][j]来的还是从dp[i][j-1]+c[i][j-1]来的,都可以在保证i != 0的情况下用c[i][j] = c[i-1][j] +/- 1 或0来算。
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
const int maxn = 5000 + 5;
const int INF = 10000000;
char p[maxn], q[maxn];
int sp[26], sq[26], ep[26], eq[26];
int d[maxn][maxn], c[maxn][maxn];
int main()
{
//freopen("D:\\txt.txt", "r", stdin);
int T, n, m;
scanf("%d", &T);
while (T--)
{
scanf("%s%s", p + 1, q + 1);
//cout << p + 1 << " " << q + 1 << endl;
n = strlen(p + 1);
m = strlen(q + 1);
//将字母转化成数字
for (int i = 1; i <= n; i++) p[i] -= 'A';
for (int i = 1; i <= m; i++) q[i] -= 'A';
//预处理
for (int i = 0; i < 26; i++)
{
sp[i] = sq[i] = INF;
ep[i] = eq[i] = 0;
}
//预处理,计算出序列1中每个字符的开始位置和结束位置
for (int i = 1; i <= n; i++)
{
sp[p[i]] = min(sp[p[i]], i);
ep[p[i]] = i;
}
//预处理序列2
for (int i = 1; i <= m; i++)
{
sq[q[i]] = min(sq[q[i]], i);
eq[q[i]] = i;
}
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= m; j++)
{
if (!i && !j) continue;
int v1 = INF, v2 = INF;
if (i) v1 = d[i-1][j] + c[i-1][j]; //从p中取颜色
if (j) v2 = d[i][j - 1] + c[i][j - 1]; //从q中取颜色
d[i][j] = min(v1, v2);
//更新c数组
if (i)
{
c[i][j] = c[i - 1][j];
if (sp[p[i]] == i && sq[p[i]] > j) c[i][j]++;
if (ep[p[i]] == i && eq[p[i]] <= j) c[i][j]--;
}
else if (j)
{
c[i][j] = c[i][j - 1];
if (sq[q[j]] == j && sp[q[j]] > i) c[i][j]++;
if (eq[q[j]] == j && ep[q[j]] <= i) c[i][j]--;
}
}
}
cout << d[n][m] << endl;
}
return 0;
}