题意：给两个字符串，然后将这两个字符串合到一起形成一个新字符串，两个字符串在新串中顺序不变，且保证新字符串长度最短，问最短长度是多少，有多少种新字符串。
题解：f[i][j]和len[i][j]分别表示str1中前i个字符和str2中前j个字符可以组成的新字符串的种类数量和长度，迭代做法，如果str1[i] == str2[j]，状态转移方程f[i][j] = f[i - 1][j - 1] len[i][j] = len[i - 1][j - 1] + 1，否则len[i][j] = min(len[i - 1][j], len[i][j - 1]), f[i][j]根据len[i - 1][j] 和 len[i][j - 1]确定。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N = 35;
char str1[N], str2[N];
int f[N][N], len[N][N], len1, len2;
int main() {
int t, cas = 1;
scanf("%d", &t);
getchar();
while (t--) {
gets(str1);
gets(str2);
len1 = strlen(str1);
len2 = strlen(str2);
for (int i = 1; i <= len1; i++) {
len[i][0] = i;
f[i][0] = 1;
}
for (int i = 1; i <= len2; i++) {
len[0][i] = i;
f[0][i] = 1;
}
f[0][0] = 1;
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (str1[i - 1] == str2[j - 1]) {
len[i][j] = len[i - 1][j - 1] + 1;
f[i][j] = f[i - 1][j - 1];
}
else {
len[i][j] = min(len[i - 1][j], len[i][j - 1]) + 1;
if (len[i - 1][j] == len[i][j - 1])
f[i][j] = f[i - 1][j] + f[i][j - 1];
else if (len[i - 1][j] > len[i][j - 1])
f[i][j] = f[i][j - 1];
else
f[i][j] = f[i - 1][j];
}
}
}
printf("Case #%d: %d %d\n", cas++, len[len1][len2], f[len1][len2]);
}
return 0;
}
