HDU 4915 Parenthese sequence

题目链接

题意：给定一个有?的左右括号串，?能替代为'('或')'，问括号匹配是否唯一或多种或不可能

思路：先从右往左扫一边，维护一个up， down表示当前位置右边右括号剩余个数的上限和下限，如果维护完后起始位置的下限为0，那么就是可以的，因为为0就代表没有多余的右括号。然后在从左往右扫一遍，和上面一样的处理，只是遇到每个问号的位置时，试一下左括号和右括号，如果都满足，表示这个位置能放左右括号，是多种可能，如果所有?都只有唯一的方法，那么答案就是唯一

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1000005;char str[N];int n, up[N], down[N], lup[N], ldown[N];bool init() {    up[n - 1] = down[n - 1] = 1;    int cnt = 0;    for (int i = n - 2; i >= 0; i--) {if (str[i] == ')') {    up[i] = up[i + 1] + 1;    down[i] = down[i + 1] + 1;}else if (str[i] == '(') {    up[i] = up[i + 1] - 1;    down[i] = down[i + 1] - 1;    if (down[i] < 0) {if (cnt == 0) return false;cnt--;if (up[i] == down[i]) up[i] = 1;down[i] = 1;    }}else {    up[i] = up[i + 1] + 1;    down[i] = down[i + 1] - 1;    if (down[i + 1] > 0 || cnt > 0) {down[i] = down[i + 1] - 1;if (down[i] < 0) {    down[i] = 1;    cnt--;}    }    else down[i] = down[i + 1] + 1;    cnt++;}    }    return (down[0] == 0);}void solve() {    n = strlen(str);        if (!init()) {printf("None\n");return;    }    lup[0] = ldown[9] = 1;    for (int i = 1; i < n - 1; i++) {if (str[i] == '(') {    lup[i] = lup[i - 1] + 1;    ldown[i] = ldown[i - 1] + 1;}else if (str[i] == ')') {    ldown[i] = ldown[i - 1] - 1;    lup[i] = lup[i - 1] - 1;    if (ldown[i] < 0) {if (lup[i] == ldown[i]) lup[i] = 1;ldown[i] = 1;    }}else {    int flag = 0;    lup[i] = lup[i - 1] + 1;    ldown[i] = ldown[i - 1] - 1;    if (ldown[i] < 0) ldown[i] = 1;    int u, d;    u = lup[i - 1] + 1;    d = ldown[i - 1] + 1;    if (u >= down[i + 1] && d <= up[i + 1])flag++;    u = max(0, lup[i - 1] - 1);    d = max(0, ldown[i - 1] - 1);    if (u >= down[i + 1] && d <= up[i + 1])flag++;    if (flag == 2) {printf("Many\n");return;    }}    }    printf("Unique\n");}int main() {    while (~scanf("%s", str)) {solve();    }    return 0;}