hdu 4545 魔法串 dp

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题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=4545

思路：设dp[i][j]为s1长度为i，s2长度为j能否成功匹配，状态出来后，详细的转移方程看代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char s1[1005], s2[1005];int m;bool g[30][30], dp[1005][1005];void solve(){int len1 = strlen(s1);int len2 = strlen(s2);memset(dp, 0, sizeof(dp));for (int i = 0; i <= len2; i++)dp[0][i] = true; //s1长度为0时肯定和s2任意长度匹配，因为只要把s2全删掉就行了for (int i = 1; i <= len1; i++){for (int j = 1; j <= len2; j++){if (dp[i][j - 1])dp[i][j] = true; //第i个字符与第j-1个可以构成，那么删除第j个即可else if (s1[i - 1] == s2[j - 1] && dp[i - 1][j - 1]) //前后两个i-1和j-1意义是不一样的，dp[i][j] = true; //当前字符相等，并且s1长度为i-1和s2长度为j-1时匹配成功，那么当前肯定匹配成功else if (g[s2[j - 1] - 'a'][s1[i - 1] - 'a'] && dp[i-1][j-1])dp[i][j] = true;//和第二种情况一样，通过给定的转换}}if (dp[len1][len2]) puts("happy");else puts("unhappy");}int main(){int t, cas = 1;scanf("%d", &t);while (t--){    scanf("%s%s", s1, s2);memset(g, false, sizeof(g));scanf("%d", &m);char u[5], v[5];for (int i = 0; i < m; i++){scanf("%s%s", u, v);g[u[0] - 'a'][v[0] - 'a'] = true;}printf("Case #%d: ", cas++);solve();}return 0;}

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