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UVaLive 7454 Parentheses (水题,贪心)

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)