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[55][55];
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 = 0; 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 = 0; i < 26; i++) changeCost[i][i] = 0;
        for (int k = 0; k < 26; k++)
            for (int i = 0; i < 26; i++)
                for (int j = 0; j < 26; 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 = 0; i < 26; i++)
            for (int j = 0; j < 26; 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 - 1; i >= 0; i--)
        {
            dp[i][i] = dp[i][i - 1] = 0;
            for (int j = i + 1; j < n; j++)
            {
                int a = word[i] - 'a', b = word[j] - 'a';
                dp[i][j] = min(dp[i][j], dp[i + 1][j] + eraseCost[a]);
                dp[i][j] = min(dp[i][j], dp[i][j - 1] + eraseCost[b]);
                for (int k = 0; k < 26; k++)
                {
                    dp[i][j] = min(dp[i][j], dp[i + 1][j] + addCost[k] + changeCost[a][k]);
                    dp[i][j] = min(dp[i][j], dp[i][j - 1] + addCost[k] + changeCost[b][k]);
                    dp[i][j] = min(dp[i][j], dp[i + 1][j - 1] + changeCost[a][k] + changeCost[b][k]);
                }
            }
        }
        LL ret = dp[0][n - 1];
        return ret >= INF ? -1 : (int)ret;
    }
};