区间DP UVA 10739 String to Palindrome

 

题目传送门

 1 /*
 2     题意：三种操作，插入，删除，替换，问最少操作数使得字符串变成回文串
 3     区间DP：有一道类似的题，有点不同的是可以替换，那么两端点不同的时候可以替换掉一个后成回文，
 4             即dp[j+1][k-1] + 1，还有这道题没有要求打印
 5 */
 6 /************************************************
 7 * Author        :Running_Time
 8 * Created Time  :2015-8-17 15:45:22
 9 * File Name     :UVA_10739.cpp
10  ************************************************/
11 
12 #include <cstdio>
13 #include <algorithm>
14 #include <iostream>
15 #include <sstream>
16 #include <cstring>
17 #include <cmath>
18 #include <string>
19 #include <vector>
20 #include <queue>
21 #include <deque>
22 #include <stack>
23 #include <list>
24 #include <map>
25 #include <set>
26 #include <bitset>
27 #include <cstdlib>
28 #include <ctime>
29 using namespace std;
30 
31 #define lson l, mid, rt << 1
32 #define rson mid + 1, r, rt << 1 | 1
33 typedef long long ll;
34 const int MAXN = 1e3 + 10;
35 const int INF = 0x3f3f3f3f;
36 const int MOD = 1e9 + 7;
37 int dp[MAXN][MAXN];
38 char str[MAXN];
39 
40 int work(void) {
41     memset (dp, 0, sizeof (dp));
42     int len = strlen (str);
43     for (int i=2; i<=len; ++i)  {
44         for (int j=0; j+i-1<len; ++j)   {
45             int k = j + i - 1;
46             int &res = dp[j][k] = INF;
47             if (str[j] == str[k])   res = dp[j+1][k-1];
48             res = min (res, min (dp[j+1][k], min (dp[j][k-1], dp[j+1][k-1])) + 1);
49         }
50     }
51     return dp[0][len-1];
52 }
53 
54 int main(void)    {     //UVA 10739 String to Palindrome
55     int T, cas = 0;  scanf ("%d", &T);
56     while (T--) {
57         scanf ("%s", str);
58         printf ("Case %d: %d\n", ++cas, work ());
59     }
60 
61     return 0;
62 }

 

转载于:https://www.cnblogs.com/Running-Time/p/4736878.html