lighto1033_区间dp

题目连接：http://lightoj.com/volume_showproblem.php?problem=1033

题目大意：给你一个字符串，问至少添加几个字符串，才能使这个字符串变成回文串

解题思路：用dp[i][j]表示[i,j]内的字符需要添加几个字符才能变回文 
考虑两种情况 
1.str[i] == str[j]，那么dp[i][j] = dp[i +1][j - 1] 
2.str[i] != str[j]，那么只能在左边添加一个，或者右边添加一个了 
所以dp[i][j] = min(dp[i + 1][j], dp[i][j - 1]) + 1

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15
16 char a[105];
17 int dp[110][110];
18 int dfs(int l, int r)
19 {
20     if(l >= r)
21         return 0;
22     if(dp[l][r] >= 0)
23         return dp[l][r];
24     int sum = INF;
25     for(int k = l+1; k <= r; k++)
26     {
27         if(a[l] == a[k])
28             sum = min(sum, dfs(l+1, k-1) + r - k);
29         else
30             sum = min(sum, min(dfs(l+1, k), dfs(l, k-1))+1+r-k);
31     }
32     dp[l][r] = sum;
33     return sum;
34 }
35 int main()
36 {
37     int t;
38     scanf("%d", &t);
39     for(int ca = 1; ca <= t; ca++)
40     {
41         scanf("%s", a);
42         int len = strlen(a);
43         memset(dp, -1, sizeof(dp));
44         int sum = dfs(0, len-1);
45
46         printf("Case %d: %d\n", ca, sum);
47     }
48     return 0;
49 }