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;
}
UVaLive 7454 Parentheses (水题，贪心)
原文：http://www.cnblogs.com/dwtfukgv/p/5877057.html