poj2955 Brackets Simple Interval DP

Tag:poj2955   brackets    simple interval dp   

poj2955 simple interval dp//d[i][j] indicates the maximum matching sequence that can be formed by the I-to-J interval//dp[i][j] = max (Dp[i][k]+dp[k+1][j]) {i<k<j}//dp[i][j] = max (dp[i+ 1][j-1]+2) if (S[i] match s[j])////memory search, will dp[i][i] = 0, the other assignment to -1;////to do the problem when the beginning will dp[i][j] into 0, the results have been tle//finally found that there are a lot of state repeated calculation , so it is not difficult to tle//this problem, learning the importance of the initial boundary! Keep practicing .... #include <algorithm> #include <bitset> #include <cassert> #include <cctype> #include <cfloat > #include <climits> #include <cmath> #include <complex> #include <cstdio> #include < cstdlib> #include <cstring> #include <ctime> #include <deque> #include <functional> #include <iostream> #include <list> #include <map> #include <numeric> #include <queue> #include <set> #include <stack> #include <vector> #define CEIL (A, B) (((a) + (b)-1)/(c)) #define Endl ' \ n ' #define GCD __gcd#define highbit (x) (1ull<< (63-__builtin_clzll (x))) #define Popcount __builtin_popcountlltypedef Long Long ll;using namespace Std;const intMOD = 1000000007;const long Double PI = ACOs ( -1.L); Template<class t> inline T LCM (const t& A, const t& b) { Return A/GCD (A, b) *b; }template<class t> Inline T lowbit (const t& x) {return x&-x;} Template<class t> inline T maximize (t& A, const t& b) {return a=a<b?b:a;} Template<class t> inline T Minimize (t& A, const t& b) {return a=a<b?a:b;} const int MAXN = 128;char s[maxn];int d[maxn][maxn];int n;void init () {memset (d,-1,sizeof (d)); n = strlen (s);} BOOL Match (char A,char b) {if (a== ' (' &&b== ') ') return true;if (a== ' [' &&b== '] ') return True;return false;  int dp (int x,int y) {if (x>=y) return d[x][y]=0;if (d[x][y]>=0) return d[x][y];//d[x][y] = 0;int& ans = D[x][y];ans  = 0;if (Match (s[x],s[y))) ans = max (ANS,DP (x+1,y-1) + 2); for (int i = x;i<y;i++) ans = max (ANS,DP (x,i) +DP (i+1,y)); return Ans;} void Solve () {printf ("%d\n", DP (0,N-1));} int main () {//freopen ("G:\\code\\1.txt", "R", stdin), while (gets (s)) {if (s[0]== ' e ')Break;init (); Solve ();} return 0;}

poj2955 Brackets Simple Interval DP