sdut 1451 括号东东 DP

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1451

题意：中文.....

思路：

pku有一道题，经典的括号匹配（区间DP）题目，那道题目是求的最长满足条件的子串的长度，那里的子串与这里的子串条件不一样。

详细：http://www.cnblogs.com/E-star/archive/2013/01/28/2879385.html

对于这个例子

)((())))(()())

pku的最长子串是12

而这里是6 

这里我们是求的连续的满足的子串。

dp[i]表示0到i的最长的满足的连续的子串

则有：

if(str[i - dp[i - 1] - 1] == '(' && str[i] == ')') dp[i] = dp[i - dp[i - 1] - 1] + 2;

if (dp[i - dp[i - 1] - 2])

dp[i] += dp[i - dp[i - 1] - 2]

//#pragma comment(linker,"/STACK:327680000,327680000")

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <functional>

#include <numeric>

#include <sstream>

#include <stack>

#include <map>

#include <queue>



#define CL(arr, val)    memset(arr, val, sizeof(arr))



#define ll long long

#define inf 0x7f7f7f7f

#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define ll long long

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)        (1 << (x))

#define iabs(x)     (x) < 0 ? -(x) : (x)

#define OUT(x)  printf("%I64d\n", x)

#define lowbit(x)   (x)&(-x)

#define Read()  freopen("din.txt", "r", stdin)

#define Write() freopen("dout.txt", "w", stdout);





#define N 1000007

using namespace std;





int dp[N];

int ans,num;

char str[N];

int n;



int main()

{

  //  Read();

    int i;

    while (~scanf("%s",str))

    {

        n = strlen(str);

        CL(dp,0);

        num = 0; ans = 0;

        for (i = 1; i < n; ++i)

        {

            if (i - dp[i - 1] - 1 >= 0 && str[i - dp[i - 1] - 1] == '(' && str[i] == ')')

            {

                dp[i] = dp[i - 1] + 2;

                if (i - dp[i - 1] - 2 >= 0 && dp[i - dp[i - 1] -2] != 0)

                {

                    dp[i] += dp[i - dp[i - 1] - 2];

                }

            }



            if (ans < dp[i])

            {

                ans = dp[i];

                num = 1;

            }

            else if (ans == dp[i]) num++;

        }

        if (ans == 0) printf("0 1\n");

        else

        printf("%d %d\n",ans,num);

    }

    return 0;



}