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SRM 509 DIV1 500pt(DP)

2024-08-04 20:05:44   225人阅读

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 };
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SRM 509 DIV1 500pt(DP)


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