HDU 4915 Parenthese sequence

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
;
int n, up
, down
, lup
, ldown
;

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;
}