UVaLive 7454 Parentheses (水题,贪心)
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题意:给定一个括号序列,改最少的括号,使得所有的括号匹配。
析:贪心,从左到右扫一下,然后统计一下左括号和右括号的数量,然后在统计中,如果有多了的右括号,那么就改成左括号,最后如果两括号数量不相等,
就改一下。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 100; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } char s[105]; int main(){ int T; cin >> T; while(T--){ scanf("%d", &n); scanf("%s", s); int ans = 0; int l = 0, r = 0; for(int i = 0; i < n; ++i){ if(s[i] == ‘)‘){ if(r == l){ ++l; ++ans; } else ++r; } else ++l; } while(l != r){ --l; ++r; ++ans; } printf("%d\n", ans); } return 0; }
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