hdu 4915 Parenthese sequence(高效)

题目链接：hdu 4915 Parenthese sequence

题目大意：给定一个序列，由(,),?组成?可以表示(或者)，问说有一种、多种或者不存在匹配。

解题思路：从左向右，从右向左，分别维护左括号和右括号可能的情况，区间上下界。如果过程中出现矛盾，则为None，否则要判断唯一解还是多解。枚举每个问号的位置，假设该问号可为左右括号，则有多解。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e6+5;

char s[maxn];
int n, l[maxn][2], r[maxn][2];

bool check (int x) {
    int ldown = l[x-1][0] + 1;
    int lup = l[x-1][1] + 1;

    if (ldown > r[x+1][1] || lup < r[x+1][0])
        return false;

    int rdown = r[x+1][0] + 1;
    int rup = r[x+1][1] + 1;

    if (rdown > l[x-1][1] || rup < l[x-1][0])
        return false;
    return true;
}

int judge () {
    n = strlen(s+1);

    if (n&1)
        return 0;

    memset(l[0], 0, sizeof(l[0]));
    memset(r[n+1], 0, sizeof(r[n+1]));

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

        if (s[i] == ‘(‘) {
            l[i][0] = l[i-1][0] + 1;
            l[i][1] = l[i-1][1] + 1;
        } else if (s[i] == ‘)‘) {

            if (l[i-1][1] == 0)
                return 0;

            l[i][0] = (l[i-1][0] == 0 ? l[i-1][0] + 2 : l[i-1][0]) - 1;
            l[i][1] = l[i-1][1] - 1;
        } else {
            l[i][0] = (l[i-1][0] == 0 ? l[i-1][0] + 2 : l[i-1][0]) - 1;
            l[i][1] = l[i-1][1] + 1;
        }
    }

    for (int i = n; i; i--) {
        if (s[i] == ‘)‘) {
            r[i][0] = r[i+1][0] + 1;
            r[i][1] = r[i+1][1] + 1;
        } else if (s[i] == ‘(‘) {

            if (r[i+1][1] == 0)
                return 0;

            r[i][0] = (r[i+1][0] == 0 ? r[i+1][0] + 2 : r[i+1][0]) - 1;
            r[i][1] = r[i+1][1] - 1;
        } else {
            r[i][0] = (r[i+1][0] == 0 ? r[i+1][0] + 2 : r[i+1][0]) - 1;
            r[i][1] = r[i+1][1] + 1;
        }
    }

    for (int i = 1; i <= n; i++)
        if (s[i] == ‘?‘ && check(i))
            return 2;
    return 1;
}

int main () {
    while (scanf("%s", s+1) == 1) {
        int flag = judge();
        if (flag == 2)
            printf("Many\n");
        else if (flag == 1)
            printf("Unique\n");
        else
            printf("None\n");
    }
    return 0;
}

hdu 4915 Parenthese sequence(高效)