SRM 509 DIV1 500pt(DP)

题目简述

给定一个字符串，可以对其进行修改，删除，增加操作，相应的操作有对应的花费，要求你用最小的花费把字符串变为回文串

题目做法

先搞一遍floyed把各种操作的最小花费求出来，然后就是类似编辑距离的DP了，这题坑了好久。。。中间结果会爆int，我设置的inf=0x3f3f3f3f,中间结果有inf+inf+inf..刚开始dp数组是int型的。。。这里搞了好几才发现这问题。。。西安现场赛也遇到这个问题了。。。然后也是浪费了将近100分钟的时间。。。导致后面做题的时间不够了。。。

代码:

 1 #define maxn 30
 2 int changeCost[maxn][maxn], addCost[maxn], eraseCost[maxn];
 3 LL dp[55][55];
 4 class PalindromizationDiv1
 5 {
 6 public:
 7     int getMinimumCost(string word, vector <string> operations)
 8     {
 9         memset(changeCost, 0x3f, sizeof(changeCost));
10         memset(addCost, 0x3f, sizeof(addCost));
11         memset(eraseCost, 0x3f, sizeof(eraseCost));
12         for (int i = 0; i < operations.size(); i++)
13         {
14             stringstream ss(operations[i]);
15             string s;
16             char a, b;
17             int num;
18             ss >> s;
19             if (s == "add")
20             {
21                 ss >> a >> num;
22                 addCost[a - ‘a‘] = num;
23             }
24             if (s == "erase")
25             {
26                 ss >> a >> num;
27                 eraseCost[a - ‘a‘] = num;
28             }
29             if (s == "change")
30             {
31                 ss >> a >> b >> num;
32                 changeCost[a - ‘a‘][b - ‘a‘] = num;
33             }
34         }
35         for (int i = 0; i < 26; i++) changeCost[i][i] = 0;
36         for (int k = 0; k < 26; k++)
37             for (int i = 0; i < 26; i++)
38                 for (int j = 0; j < 26; j++)
39                 {
40                     if (i == j || j == k || i == k) continue;
41                     changeCost[i][j] = min(changeCost[i][j], changeCost[i][k] + changeCost[k][j]);
42                 }
43
44         for (int i = 0; i < 26; i++)
45             for (int j = 0; j < 26; j++)
46             {
47                 addCost[i] = min(addCost[i], addCost[j] + changeCost[j][i]);
48                 eraseCost[i] = min(eraseCost[i], changeCost[i][j] + eraseCost[j]);
49             }
50         int n = word.size();
51         memset(dp, 0x3f, sizeof(dp));
52         for (int i = n - 1; i >= 0; i--)
53         {
54             dp[i][i] = dp[i][i - 1] = 0;
55             for (int j = i + 1; j < n; j++)
56             {
57                 int a = word[i] - ‘a‘, b = word[j] - ‘a‘;
58                 dp[i][j] = min(dp[i][j], dp[i + 1][j] + eraseCost[a]);
59                 dp[i][j] = min(dp[i][j], dp[i][j - 1] + eraseCost[b]);
60                 for (int k = 0; k < 26; k++)
61                 {
62                     dp[i][j] = min(dp[i][j], dp[i + 1][j] + addCost[k] + changeCost[a][k]);
63                     dp[i][j] = min(dp[i][j], dp[i][j - 1] + addCost[k] + changeCost[b][k]);
64                     dp[i][j] = min(dp[i][j], dp[i + 1][j - 1] + changeCost[a][k] + changeCost[b][k]);
65                 }
66             }
67         }
68         LL ret = dp[0][n - 1];
69         return ret >= INF ? -1 : (int)ret;
70     }
71 };