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题目大意：给出三个字符串，从分别从第一个字符串和第二个字符串中挑选子串a,b，用a和b组成第三个字符串，问可组成的子串有多少种。

解题思路：dp[i][j][k]表是用s1中的前i个字符和s2中的前j个字符的子串组成s3前k个字符的情况。

仿照http://www.cnblogs.com/yuzhaoxin/archive/2012/05/04/2483259.html

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int DP[45][45][45][45], A[4][45], top[4], vis[180], n;int DPS(int cur) {int &ans = DP[top[0]][top[1]][top[2]][top[3]];if (ans != -1)return ans;if (cur == 5)return 0;int maxa = 0;for (int i = 0; i < 4; i++) { if (top[i] == n)continue;int color = A[i][top[i]];top[i]++;if (vis[color]) {vis[color] = 0;maxa = max(maxa, DPS(cur-1) + 1);vis[color] = 1;}else {vis[color] = 1;maxa = max(maxa, DPS(cur+1));vis[color] = 0;}top[i]--;}return ans = maxa;}int main() {while (scanf("%d", &n), n) {for (int i = 0; i < n; i++)for (int j = 0; j < 4; j++)scanf("%d", &A[j][i]);memset(DP, -1, sizeof(DP));memset(top, 0, sizeof(top));memset(vis, 0, sizeof(vis));printf("%d\n", DPS(0));}return 0;}

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