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:

 #define maxn 30
 int changeCost[maxn][maxn], addCost[maxn], eraseCost[maxn];
 LL dp[][];
 class PalindromizationDiv1
 {
 public:
     int getMinimumCost(string word, vector <string> operations)
     {
         memset(changeCost, 0x3f, sizeof(changeCost));
         memset(addCost, 0x3f, sizeof(addCost));
         memset(eraseCost, 0x3f, sizeof(eraseCost));
         for (int i = ; i < operations.size(); i++)
         {
             stringstream ss(operations[i]);
             string s;
             char a, b;
             int num;
             ss >> s;
             if (s == "add")
             {
                 ss >> a >> num;
                 addCost[a - 'a'] = num;
             }
             if (s == "erase")
             {
                 ss >> a >> num;
                 eraseCost[a - 'a'] = num;
             }
             if (s == "change")
             {
                 ss >> a >> b >> num;
                 changeCost[a - 'a'][b - 'a'] = num;
             }
         }
         for (int i = ; i < ; i++) changeCost[i][i] = ;
         for (int k = ; k < ; k++)
             for (int i = ; i < ; i++)
                 for (int j = ; j < ; j++)
                 {
                     if (i == j || j == k || i == k) continue;
                     changeCost[i][j] = min(changeCost[i][j], changeCost[i][k] + changeCost[k][j]);
                 }

         for (int i = ; i < ; i++)
             for (int j = ; j < ; j++)
             {
                 addCost[i] = min(addCost[i], addCost[j] + changeCost[j][i]);
                 eraseCost[i] = min(eraseCost[i], changeCost[i][j] + eraseCost[j]);
             }
         int n = word.size();
         memset(dp, 0x3f, sizeof(dp));
         for (int i = n - ; i >= ; i--)
         {
             dp[i][i] = dp[i][i - ] = ;
             for (int j = i + ; j < n; j++)
             {
                 int a = word[i] - 'a', b = word[j] - 'a';
                 dp[i][j] = min(dp[i][j], dp[i + ][j] + eraseCost[a]);
                 dp[i][j] = min(dp[i][j], dp[i][j - ] + eraseCost[b]);
                 for (int k = ; k < ; k++)
                 {
                     dp[i][j] = min(dp[i][j], dp[i + ][j] + addCost[k] + changeCost[a][k]);
                     dp[i][j] = min(dp[i][j], dp[i][j - ] + addCost[k] + changeCost[b][k]);
                     dp[i][j] = min(dp[i][j], dp[i + ][j - ] + changeCost[a][k] + changeCost[b][k]);
                 }
             }
         }
         LL ret = dp[][n - ];
         return ret >= INF ? - : (int)ret;
     }
 };