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 
   
     #include 
    
      #include 
     
       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; }