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;
}