首页 > > 程序设计 > C/C++ >

HDU6299 Balanced Sequence (多校第一场1002) …

2018-09-10 00:58:22来源：博客园阅读 ()

Balanced Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6207 Accepted Submission(s): 1616

Problem Description

Chiaki has

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test case, output an integer denoting the answer.

Sample Input

2 1 )()(()( 2 ) )(

Sample Output

4 2

题目意思就是给出几组括号序列，然后排序，使得匹配的括号数最大。

首先先处理每一串序列，最后的形式都是))) (((( , 然后就是贪心排序四种情况。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <string.h>
 5 #include <stdlib.h>
 6 #include <string>
 7 #include <vector>
 8 #include <set>
 9 #include <map>
10 #include <queue>
11 #include <algorithm>
12 #include <sstream>
13 #include <stack>
14 using namespace std;
15 #define rep(i,a,n) for (int i=a;i<n;i++)
16 #define per(i,a,n) for (int i=n-1;i>=a;i--)
17 #define pb push_back
18 #define mp make_pair
19 #define all(x) (x).begin(),(x).end()
20 #define fi first
21 #define se second
22 #define SZ(x) ((int)(x).size())
23 #define FO freopen("in.txt", "r", stdin)
24 #define lowbit(x) (x&-x)
25 #define mem(a,b) memset(a, b, sizeof(a))
26 typedef vector<int> VI;
27 typedef long long ll;
28 typedef pair<int,int> PII;
29 const ll mod=1000000007;
30 const int inf = 0x3f3f3f3f;
31 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
32 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
33 //head
34 
35 const int maxn = 100010;
36 int _, n;
37 struct node{
38     int l, r, ans;// 左， 右， 单匹配
39     bool operator < (node &x) const {//排序很重要
40         if(r >= l && x.r < x.l) return false; //右括号少的排前面
41         if(r < l && x.r >= x.l) return true;
42         if(r >= l && x.r >= x.l) return l > x.l ; //都是右括号大于左括号 左括号多的排前面
43         return r < x.r; // 都是左括号大于右括号 右括号少的排前面
44     }
45 }a[maxn];
46 
47 char s[maxn];
48 void solve() {
49     scanf("%d", &n);
50     rep(i, 0, n) {
51         a[i].l = a[i].r = a[i].ans = 0;
52         scanf("%s", s);
53         int len = strlen(s);
54         rep(j, 0, len) {
55             if(s[j] == '(')
56                 a[i].l++;
57             else {
58                 if(a[i].l > 0)
59                     a[i].l--, a[i].ans++;
60                 else
61                     a[i].r++;
62             }
63         }
64     }
65     sort(a, a+n);
66     int now = 0;//维护 (
67     int sum = 0;
68     rep(i, 0, n) {
69         if(a[i].r > now)//如果 ‘）’ 大于 ‘（’ 
70             a[i].r = now; //最多匹配 的是 ‘（’  若不大于的话，就不用变
71         sum += a[i].r + a[i].ans;//然后累加
72         now -= a[i].r;//更新 当前的  ‘（’ 数量
73         now += a[i].l;
74     }
75     printf("%d\n", 2 * sum);
76 }
77 
78 int main() {
79     for(scanf("%d", &_);_;_--) {
80         solve();
81     }
82 }