3D -- minimum cost input string

Give you a sequence ),(,? Composition, number I? The cost is different from that of the right bracket. Now let you form a back-to-text sequence and make it the least costly.

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <stack>#include <map>#include <queue>using namespace std;struct cost {    int p;    int v;    bool operator < (const cost& a) const {        return v < a.v;    }};cost make(int p, int v) {    cost ret;    ret.p = p;    ret.v = v;    return ret;}char str[50004];int main() {    scanf("%s", str);    int cnt = 0;    int len = strlen(str);    int a, b;    priority_queue<cost> q;    long long ans = 0;    for (int i = 0; i < len; i++) {        cnt += str[i] == ‘(‘;        cnt -= str[i] == ‘)‘ || str[i] == ‘?‘;        if (str[i] == ‘?‘) {            scanf("%d %d", &a, &b);            q.push(make(i, b-a));            ans += b;            str[i] = ‘)‘;        }        if (cnt < 0 && q.empty()) {            ans = -1;            break;        }        if (cnt < 0) {            cost top = q.top();            q.pop();            ans = ans - top.v;            str[top.p] = ‘(‘;            cnt += 2;        }    }    if (cnt > 0) ans = -1;    printf("%I64d\n", ans);    if (ans != -1) printf("%s\n", str);    return 0;}

 

3D -- minimum cost input string