题目链接：http://poj.org/problem?id=1159
题意：求一个字符串加多少个字符，可以变成一个回文串。
把这个字符串倒过来存一遍，求这两个字符串的lcs，用原长减去lcs就行。这题卡内存真稀奇，于是修改成滚动数组。观察发现i值的更新只有可能是从i或i-1转移来，所以就i取模2。
1 /* 2 ━━━━━┒ギリギリ♂ eye！ 3 ┓┏┓┏┓┃キリキリ♂ mind！ 4 ┛┗┛┗┛┃＼○／ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rll(a) scanf("%lld", &a) 44 #define Rs(a) scanf("%s", a) 45 #define Cin(a) cin >> a 46 #define FRead() freopen("in", "r", stdin) 47 #define FWrite() freopen("out", "w", stdout) 48 #define Rep(i, len) for(int i = 0; i < (len); i++) 49 #define For(i, a, len) for(int i = (a); i < (len); i++) 50 #define Cls(a) memset((a), 0, sizeof(a)) 51 #define Clr(a, x) memset((a), (x), sizeof(a)) 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a)) 53 #define lrt rt << 1 54 #define rrt rt << 1 | 1 55 #define pi 3.14159265359 56 #define RT return 57 #define lowbit(x) x & (-x) 58 #define onenum(x) __builtin_popcount(x) 59 typedef long long LL; 60 typedef long double LD; 61 typedef unsigned long long ULL; 62 typedef pair<int, int> pii; 63 typedef pair<string, int> psi; 64 typedef pair<LL, LL> pll; 65 typedef map<string, int> msi; 66 typedef vector<int> vi; 67 typedef vector<LL> vl; 68 typedef vector<vl> vvl; 69 typedef vector<bool> vb; 70 71 const int maxn = 5050; 72 int n; 73 int dp[2][maxn]; 74 char s[maxn]; 75 char t[maxn]; 76 77 int main() { 78 // FRead(); 79 while(~Rint(n)) { 80 Cls(t); Cls(dp); 81 Rs(s+1); 82 For(i, 1, n+1) t[n-i+1] = s[i]; 83 For(i, 1, n+1) { 84 For(j, 1, n+1) { 85 if(s[i] == t[j]) dp[i%2][j] = max(dp[i%2][j], dp[(i-1)%2][j-1]+1); 86 dp[i%2][j] = max(dp[i%2][j], max(dp[(i-1)%2][j], dp[i%2][j-1])); 87 } 88 } 89 printf("%d ", n - dp[n%2][n]); 90 } 91 RT 0; 92 }