Hdu 4915 Parenthese sequence (efficient)

Hdu 4915 Parenthese sequence (efficient)

Link: hdu 4915 Parenthese sequence

Question: Given a sequence, (,),? Composition? It can represent (OR), and there is one, multiple, or no matching.

Solution: Maintain the possible conditions of left and right brackets from left to right and the upper and lower bounds of the range respectively. If there is a conflict in the process, it is None. Otherwise, you must determine whether the unique solution is a multiple solution. Enumerate the positions of each question mark. If the question mark can be left or right parentheses, multiple solutions are provided.

#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;}