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;

}