UVA1625Color Lenth(DP+LCS变形 未AC）

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=105116#problem/C

紫书P276

res[i][j]表示第一个序列移动i个，第二个序列移动j个之后有几个已经出现但尚未结束,dp[i][j]表示第一个序列移动i个，第二个序列移动j个之后的总长度

dp[i][j] = min(dp[i - 1][j],dp[i][j - 1]) + res[i][j]

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 const int MAX =  + ;
 const int INF = 0x3f3f3f3f;
 char str1[MAX],str2[MAX];
 int dp[MAX][MAX],res[MAX][MAX];
 int start1[],End1[],start2[],End2[],vis[];
 int n,m;
 int main()
 {
     int tase;
     scanf("%d", &tase);
     while(tase--)
     {
         scanf("%s%s", str1 + ,str2 + );
         n = strlen(str1 + );
         m = strlen(str2 + );
         memset(vis, , sizeof(vis));
         memset(End1, -, sizeof(End1));
         memset(End2, -, sizeof(End2));
         for(int i = ; i < ; i++)
             start1[i] = start2[i] = INF;
         for(int i = ; i <= n; i++)
         {
             if(start1[str1[i] - 'A'] == INF)
                 start1[str1[i] - 'A'] = i;
             End1[str1[i] - 'A'] = i;
         }
         for(int i = ; i <= m; i++)
         {
             if(start2[str2[i] - 'A'] == INF)
                 start2[str2[i] - 'A'] = i ;
             End2[str2[i] - 'A'] = i;
         }
 
         memset(res, , sizeof(res));
         for(int i = ; i <= n; i++)
         {
             for(int j = ; j <= m; j++)
             {
                 int cnt = ;
                 for(int k = ; k < ; k++)
                 {
                     if(start1[k] == INF && start2[k] == INF)
                         continue;
                     if(start1[k] > i && start2[k] > j)
                         continue;
                     if(End1[k] <= i && End2[k] <= j)
                         continue;
                     cnt++;
                 }
                 res[i][j] = cnt;
             }
         }
 
         dp[][] = ;
         for(int i = ; i <= m ; i++)
             dp[][i] = dp[][i - ] + res[][i];
         for(int j = ; j <= n; j++)
             dp[j][] = dp[j - ][] + res[j][];
         for(int i = ; i <= n; i++)
         {
             for(int j = ; j <= m; j++)
             {
                 dp[i][j] = min(dp[i - ][j], dp[i][j - ]) + res[i][j];
             }
         }
 
         printf("%d\n", dp[n][m]);
 
     }
 
     return ;
 }

TLE