[HIHO1323]回文字符串（区间dp）

题目链接：http://hihocoder.com/problemset/problem/1323

思路：区间dp，按照区间长度枚举所有区间和区间的起始位置。这时也可获取到区间的末位，比对这两个字符是否相同，如果相同可以更新dp(i,j)=max(dp(i+1,j-1), dp(i,j))。否则，更新三种操作：
dp(i,j)=min(dp(i+1,j),dp(i,j-1),dp(i+1,j-1))+1

     /*
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 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
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 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;

 int dp[maxn][maxn];
 char s[maxn];
 int n;

 int main() {
     // FRead();
     while(~Rs(s+)) {
         n = strlen(s+);
         Cls(dp);
         For(k, , n+) {
             For(i, , n-k+) {
                 int j = i + k;
                 dp[i][j] = n;
                 if(s[i] == s[j]) dp[i][j] = min(dp[i][j], dp[i+][j-]);
                 dp[i][j] = min(dp[i+][j]+, dp[i][j]);
                 dp[i][j] = min(dp[i][j-]+, dp[i][j]);
                 dp[i][j] = min(dp[i+][j-]+, dp[i][j]);
             }
         }
         printf("%d\n", dp[][n]);
     }
     RT ;
 }