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POJ 2955 Brackets (区间DP)

题意:给定一个序列,问你最多有多少个合法的括号。

析:区间DP,dp[i][j] 表示在 第 i 到 第 j 区间内最多有多少个合法的括号。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <string>#include <cstdlib>#include <cmath>#include <iostream>#include <cstring>#include <set>#include <queue>#include <algorithm>#include <vector>#include <map>#include <cctype>#include <cmath>#include <stack>//#include <tr1/unordered_map>#define freopenr freopen("in.txt", "r", stdin)#define freopenw freopen("out.txt", "w", stdout)using namespace std;//using namespace std :: tr1;typedef long long LL;typedef pair<int, int> P;const int INF = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f;const LL LNF = 0x3f3f3f3f3f3f;const double PI = acos(-1.0);const double eps = 1e-8;const int maxn = 1e2 + 5;const int mod = 1e9 + 7;const int N = 1e6 + 5;const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }int n, m;const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};inline int Min(int a, int b){ return a < b ? a : b; }inline int Max(int a, int b){ return a > b ? a : b; }inline LL Min(LL a, LL b){ return a < b ? a : b; }inline LL Max(LL a, LL b){ return a > b ? a : b; }inline bool is_in(int r, int c){    return r >= 0 && r < n && c >= 0 && c < m;}char s[maxn];int dp[maxn][maxn];bool match(char ch, char sh){    if(ch == ‘(‘ && sh == ‘)‘)  return true;    if(ch == ‘[‘ && sh == ‘]‘)  return true;    return false;}int main(){    while(scanf("%s", s) == 1 && s[0] != ‘e‘){        n = strlen(s);        memset(dp, 0, sizeof dp);        for(int i = n-2; i >= 0; --i)            for(int j = i+1; j < n; ++j){                if(match(s[i], s[j]))  dp[i][j] = dp[i+1][j-1] + 2;                for(int k = i; k < j; ++k)                    dp[i][j] = Max(dp[i][k]+dp[k+1][j], dp[i][j]);            }        printf("%d\n", dp[0][n-1]);    }    return 0;}

 

POJ 2955 Brackets (区间DP)