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]

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 const int MAX = 5000 + 10;
 7 const int INF = 0x3f3f3f3f;
 8 char str1[MAX],str2[MAX];
 9 int dp[MAX][MAX],res[MAX][MAX];
10 int start1[30],End1[30],start2[30],End2[30],vis[30];
11 int n,m;
12 int main()
13 {
14     int tase;
15     scanf("%d", &tase);
16     while(tase--)
17     {
18         scanf("%s%s", str1 + 1,str2 + 1);
19         n = strlen(str1 + 1);
20         m = strlen(str2 + 1);
21         memset(vis, 0, sizeof(vis));
22         memset(End1, -1, sizeof(End1));
23         memset(End2, -1, sizeof(End2));
24         for(int i = 0; i < 27; i++)
25             start1[i] = start2[i] = INF;
26         for(int i = 1; i <= n; i++)
27         {
28             if(start1[str1[i] - 'A'] == INF)
29                 start1[str1[i] - 'A'] = i;
30             End1[str1[i] - 'A'] = i;
31         }
32         for(int i = 1; i <= m; i++)
33         {
34             if(start2[str2[i] - 'A'] == INF)
35                 start2[str2[i] - 'A'] = i ;
36             End2[str2[i] - 'A'] = i;
37         }
38 
39         memset(res, 0, sizeof(res));
40         for(int i = 0; i <= n; i++)
41         {
42             for(int j = 0; j <= m; j++)
43             {
44                 int cnt = 0;
45                 for(int k = 0; k < 26; k++)
46                 {
47                     if(start1[k] == INF && start2[k] == INF)
48                         continue;
49                     if(start1[k] > i && start2[k] > j)
50                         continue;
51                     if(End1[k] <= i && End2[k] <= j)
52                         continue;
53                     cnt++;
54                 }
55                 res[i][j] = cnt;
56             }
57         }
58         
59         dp[0][0] = 0;
60         for(int i = 1; i <= m ; i++)
61             dp[0][i] = dp[0][i - 1] + res[0][i];
62         for(int j = 1; j <= n; j++)
63             dp[j][0] = dp[j - 1][0] + res[j][0];
64         for(int i = 1; i <= n; i++)
65         {
66             for(int j = 1; j <= m; j++)
67             {
68                 dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + res[i][j];
69             }
70         }
71 
72         printf("%d\n", dp[n][m]);
73 
74     }
75 
76     return 0;
77 }

TLE