【POJ 2955】Brackets（区间DP）

【POJ 2955】Brackets（区间DP）

入门级区间DP

问最长的匹配括号长度。只包含()和[]
要求匹配括号不可交叉，即([)]这种不计入

因为不计交叉情况，转移就很直白。

枚举区间长度l，转移为

if( (str[i] == '(' && str[i+l-1] == ')') || (str[i] == '[' && str[i+l-1] == ']') )     dp[i][i+l-1] = max(dp[i][i+l-1],dp[i+1][i+l-2]+2);//合并相邻区间最长括号匹配数for(int j = i; j < i+l; ++j)    dp[i][i+l-1] = max(dp[i][i+l-1],dp[i][j]+dp[j+1][i+l-1]);

代码如下：

#include <iostream>#include <cmath>#include <vector>#include <cstdlib>#include <cstdio>#include <climits>#include <ctime>#include <cstring>#include <queue>#include <stack>#include <list>#include <algorithm>#include <map>#include <set>#define LL long long#define Pr pair<int,int>#define fread(ch) freopen(ch,"r",stdin)#define fwrite(ch) freopen(ch,"w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int msz = 10000;const int mod = 1e9+7;const double eps = 1e-8;char str[233];int dp[233][233];bool cal(int i,int j){    return (str[i] == '(' && str[j] == ')') || (str[i] == '[' && str[j] == ']');}int main(){    //fread("");    //fwrite("");    while(~scanf("%s",str) && str[0] != 'e')    {        int len = strlen(str);        memset(dp,0,sizeof(dp));        for(int l = 2; l <= len; l++)        {            for(int i = 0; i+l <= len; ++i)            {                if(cal(i,i+l-1)) dp[i][i+l-1] = max(dp[i][i+l-1],dp[i+1][i+l-2]+2);                for(int j = i; j < i+l; ++j)                    dp[i][i+l-1] = max(dp[i][i+l-1],dp[i][j]+dp[j+1][i+l-1]);            }        }        printf("%d\n",dp[0][len-1]);    }    return 0;}